Physics-
General
Easy

Question

A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction ‘m’ A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

  1. mg    
  2. fraction numerator m g over denominator 4 end fraction    
  3. fraction numerator m g over denominator 2 end fraction    
  4. m g left parenthesis 1 minus mu right parenthesis    

The correct answer is: fraction numerator m g over denominator 2 end fraction


    applying the condition of rotational equilibrium
    F left parenthesis L right parenthesis equals m g open square brackets fraction numerator L over denominator 2 end fraction close square brackets semicolon rightwards double arrow F equals fraction numerator m g over denominator 2 end fraction

    Related Questions to study

    General
    physics-

    If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of intersection of l-T and l minus T to the power of 2 end exponent graph

    T equals 2 pi square root of fraction numerator l over denominator g end fraction end root semicolon T to the power of 2 end exponent equals 4 pi to the power of 2 end exponent fraction numerator l over denominator g end fraction
    l equals fraction numerator g over denominator 4 pi to the power of 2 end exponent end fraction equals fraction numerator 9.8 over denominator 4 cross times 9.8 end fraction equals 0.25 equals 25 c m

    If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of intersection of l-T and l minus T to the power of 2 end exponent graph

    physics-General
    T equals 2 pi square root of fraction numerator l over denominator g end fraction end root semicolon T to the power of 2 end exponent equals 4 pi to the power of 2 end exponent fraction numerator l over denominator g end fraction
    l equals fraction numerator g over denominator 4 pi to the power of 2 end exponent end fraction equals fraction numerator 9.8 over denominator 4 cross times 9.8 end fraction equals 0.25 equals 25 c m
    General
    Maths-

    Order and degree of differential equation fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent are

    fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 4 equals y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 4 minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared equals y
N o w space w e space c a n space s e e space t h a t space o r d e r space equals 2 space a n d space d e g r e e equals 4

    Order and degree of differential equation fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent are

    Maths-General
    fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 4 equals y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 4 minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared equals y
N o w space w e space c a n space s e e space t h a t space o r d e r space equals 2 space a n d space d e g r e e equals 4
    General
    Maths-

    The solution of x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x is

    x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x
rightwards double arrow x squared fraction numerator d y over denominator d x end fraction minus x y equals 2 cos squared fraction numerator y over denominator 2 x end fraction
rightwards double arrow fraction numerator 1 over denominator cos squared begin display style fraction numerator y over denominator 2 x end fraction end style end fraction open square brackets x squared. fraction numerator d y over denominator d x end fraction minus x y close square brackets equals 2
rightwards double arrow 1 half s e c squared fraction numerator y over denominator 2 x end fraction open square brackets fraction numerator x. begin display style fraction numerator d y over denominator d x end fraction end style minus y over denominator x squared end fraction close square brackets equals 1 over x cubed
rightwards double arrow fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction equals 1 over x cubed
O n space i n t e g r a t i n g space w e space g e comma
integral fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction d x equals integral 1 over x cubed d x
tan fraction numerator y over denominator 2 x end fraction equals C minus fraction numerator 1 over denominator 2 x squared end fraction

    The solution of x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x is

    Maths-General
    x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x
rightwards double arrow x squared fraction numerator d y over denominator d x end fraction minus x y equals 2 cos squared fraction numerator y over denominator 2 x end fraction
rightwards double arrow fraction numerator 1 over denominator cos squared begin display style fraction numerator y over denominator 2 x end fraction end style end fraction open square brackets x squared. fraction numerator d y over denominator d x end fraction minus x y close square brackets equals 2
rightwards double arrow 1 half s e c squared fraction numerator y over denominator 2 x end fraction open square brackets fraction numerator x. begin display style fraction numerator d y over denominator d x end fraction end style minus y over denominator x squared end fraction close square brackets equals 1 over x cubed
rightwards double arrow fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction equals 1 over x cubed
O n space i n t e g r a t i n g space w e space g e comma
integral fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction d x equals integral 1 over x cubed d x
tan fraction numerator y over denominator 2 x end fraction equals C minus fraction numerator 1 over denominator 2 x squared end fraction
    parallel
    General
    physics-

    For the given figure, calculate zero correction.

    For the given figure, calculate zero correction.

    physics-General
    General
    physics-

    Three capacitors C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript are connected as shown in the figure to a battery of V volt. If the capacitor C subscript 3 end subscriptbreaks down electrically the change in total charge on the combination of capacitors is

    Since,C subscript 1 end subscript a n d blank C subscript 2 end subscript are parallel to their equivalent capacitance will be left parenthesis C subscript 1 end subscript plus C subscript 2 end subscript right parenthesis. Now, open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses a n d blank C subscript 3 end subscript are in series, so the net equivalent capacitances of circuit.
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C to the power of 3 end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction
    equals blank fraction numerator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript over denominator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript end fraction
    C equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    Since, V is the voltage of the battery, so charge on this system
    q equals C V
    q equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    If the capacitor C subscript 3 end subscript breaks down then total equivalent capacitance
    C to the power of ´ end exponent equals blank C subscript 1 end subscript plus C subscript 2 end subscript
    therefore New charge stored
    q to the power of ´ end exponent equals C ´ V
    q ´ equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V
    Change in total charge
    increment q equals q to the power of ´ end exponent minus q left parenthesis because q to the power of ´ end exponent greater than q right parenthesis
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V minus fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets
    increment q equals blank open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets

    Three capacitors C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript are connected as shown in the figure to a battery of V volt. If the capacitor C subscript 3 end subscriptbreaks down electrically the change in total charge on the combination of capacitors is

    physics-General
    Since,C subscript 1 end subscript a n d blank C subscript 2 end subscript are parallel to their equivalent capacitance will be left parenthesis C subscript 1 end subscript plus C subscript 2 end subscript right parenthesis. Now, open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses a n d blank C subscript 3 end subscript are in series, so the net equivalent capacitances of circuit.
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C to the power of 3 end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction
    equals blank fraction numerator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript over denominator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript end fraction
    C equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    Since, V is the voltage of the battery, so charge on this system
    q equals C V
    q equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    If the capacitor C subscript 3 end subscript breaks down then total equivalent capacitance
    C to the power of ´ end exponent equals blank C subscript 1 end subscript plus C subscript 2 end subscript
    therefore New charge stored
    q to the power of ´ end exponent equals C ´ V
    q ´ equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V
    Change in total charge
    increment q equals q to the power of ´ end exponent minus q left parenthesis because q to the power of ´ end exponent greater than q right parenthesis
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V minus fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets
    increment q equals blank open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets
    General
    physics-

    The charge deposited on 4 mu F capacitor the circuit is


    As the capacitors 4 mu F and 2 mu F are connected in parallel and are in series with 6blank mu F capacitor, their equivalent capacitance is
    fraction numerator left parenthesis 2 plus 4 right parenthesis cross times 6 over denominator 2 plus 4 plus 6 end fraction equals 3 blank mu F
    Charge in the circuit
    Q equals blank 3 blank mu F cross times 12 V equals 36 mu C

    Since, the capacitors 4 mu F and 2mu F are connected in parallel, therefore potential difference across them is same.
    rightwards double arrow fraction numerator Q subscript 1 end subscript over denominator Q subscript 2 end subscript end fraction equals fraction numerator C subscript 1 end subscript over denominator C subscript 2 end subscript end fraction equals fraction numerator 4 over denominator 2 end fraction o r Q subscript 1 end subscript equals 2 Q subscript 2 end subscript
    Also Q equals Q subscript 1 end subscript plus Q subscript 2 end subscript
    therefore blank 36 mu C equals 2 Q subscript 2 end subscript plus Q subscript 2 end subscript or Q subscript 2 end subscript equals fraction numerator 36 mu C over denominator 3 end fraction equals 12 mu C
    Q subscript 1 end subscript equals Q minus Q subscript 2 end subscript equals 36 mu C minus 12 mu C
    equals 24 blank mu C equals 24 cross times 10 to the power of negative 6 end exponent C

    The charge deposited on 4 mu F capacitor the circuit is

    physics-General

    As the capacitors 4 mu F and 2 mu F are connected in parallel and are in series with 6blank mu F capacitor, their equivalent capacitance is
    fraction numerator left parenthesis 2 plus 4 right parenthesis cross times 6 over denominator 2 plus 4 plus 6 end fraction equals 3 blank mu F
    Charge in the circuit
    Q equals blank 3 blank mu F cross times 12 V equals 36 mu C

    Since, the capacitors 4 mu F and 2mu F are connected in parallel, therefore potential difference across them is same.
    rightwards double arrow fraction numerator Q subscript 1 end subscript over denominator Q subscript 2 end subscript end fraction equals fraction numerator C subscript 1 end subscript over denominator C subscript 2 end subscript end fraction equals fraction numerator 4 over denominator 2 end fraction o r Q subscript 1 end subscript equals 2 Q subscript 2 end subscript
    Also Q equals Q subscript 1 end subscript plus Q subscript 2 end subscript
    therefore blank 36 mu C equals 2 Q subscript 2 end subscript plus Q subscript 2 end subscript or Q subscript 2 end subscript equals fraction numerator 36 mu C over denominator 3 end fraction equals 12 mu C
    Q subscript 1 end subscript equals Q minus Q subscript 2 end subscript equals 36 mu C minus 12 mu C
    equals 24 blank mu C equals 24 cross times 10 to the power of negative 6 end exponent C
    parallel
    General
    Maths-

    If y equals e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis then y satisfies

    I f space y equals e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis
d i f f e r e n t i a t i n g space b o t h space s i d e s comma space w e space g e t
fraction numerator d y over denominator d x end fraction equals fraction numerator d over denominator d x end fraction e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis
rightwards double arrow fraction numerator d y over denominator d x end fraction equals negative e to the power of negative x end exponent open parentheses A cos x plus B sin x close parentheses plus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses
rightwards double arrow fraction numerator d y over denominator d x end fraction equals negative y plus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses
A g a i n comma space d i f f e r e n t i a t i n g space b o t h space s i d e s comma space w e space g e t
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator negative d y over denominator d x end fraction plus e to the power of negative x end exponent open parentheses negative A cos x minus B sin x close parentheses minus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction equals negative e to the power of negative x end exponent open parentheses A cos x plus B sin x close parentheses minus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction equals negative y minus y minus fraction numerator d y over denominator d x end fraction
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction plus 2 y equals 0

    If y equals e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis then y satisfies

    Maths-General
    I f space y equals e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis
d i f f e r e n t i a t i n g space b o t h space s i d e s comma space w e space g e t
fraction numerator d y over denominator d x end fraction equals fraction numerator d over denominator d x end fraction e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis
rightwards double arrow fraction numerator d y over denominator d x end fraction equals negative e to the power of negative x end exponent open parentheses A cos x plus B sin x close parentheses plus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses
rightwards double arrow fraction numerator d y over denominator d x end fraction equals negative y plus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses
A g a i n comma space d i f f e r e n t i a t i n g space b o t h space s i d e s comma space w e space g e t
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator negative d y over denominator d x end fraction plus e to the power of negative x end exponent open parentheses negative A cos x minus B sin x close parentheses minus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction equals negative e to the power of negative x end exponent open parentheses A cos x plus B sin x close parentheses minus e to the power of negative x end exponent open parentheses negative A sin x plus B cos x close parentheses
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction equals negative y minus y minus fraction numerator d y over denominator d x end fraction
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction plus 2 y equals 0
    General
    physics-

    In the figure, a proton moves a distance d in a uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease?

    Since, the proton is moving against the direction of electric field so, work is done by the proton against electric field. It implies that electric field does negative work on the proton.
    Again, proton is moving in electric field from low potential region to high potential region hence, its potential energy increases.

    In the figure, a proton moves a distance d in a uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease?

    physics-General
    Since, the proton is moving against the direction of electric field so, work is done by the proton against electric field. It implies that electric field does negative work on the proton.
    Again, proton is moving in electric field from low potential region to high potential region hence, its potential energy increases.
    General
    maths-

    Solution of fraction numerator d y over denominator d x end fraction equals e to the power of y plus x end exponent plus e to the power of y minus x end exponent is

    Solution of fraction numerator d y over denominator d x end fraction equals e to the power of y plus x end exponent plus e to the power of y minus x end exponent is

    maths-General
    parallel
    General
    physics-

    Two charges q subscript 1 end subscriptand q subscript 2 end subscript are placed 30 cm apart, as shown in the figure. A third charge q subscript 3 end subscriptis moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k comma where k is

    When charge q subscript 3 end subscriptis at C, then its potential energy is
    U subscript C end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    Where charge q subscript 3 end subscriptis at D, then
    U subscript D end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction close parentheses
    Hence, change in potential energy
    increment U equals U subscript D end subscript minus U subscript C end subscript
    equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    b u t blank increment U equals fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k
    therefore fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    ⟹ k equals q subscript 2 end subscript open parentheses 10 minus 2 close parentheses equals 8 q subscript 2 end subscript

    Two charges q subscript 1 end subscriptand q subscript 2 end subscript are placed 30 cm apart, as shown in the figure. A third charge q subscript 3 end subscriptis moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k comma where k is

    physics-General
    When charge q subscript 3 end subscriptis at C, then its potential energy is
    U subscript C end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    Where charge q subscript 3 end subscriptis at D, then
    U subscript D end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction close parentheses
    Hence, change in potential energy
    increment U equals U subscript D end subscript minus U subscript C end subscript
    equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    b u t blank increment U equals fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k
    therefore fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    ⟹ k equals q subscript 2 end subscript open parentheses 10 minus 2 close parentheses equals 8 q subscript 2 end subscript
    General
    physics-

    Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential differenceV subscript A end subscript minus V subscript B end subscript equals V subscript B end subscript minus V subscript C end subscript. If t subscript 1 end subscriptand t subscript 2 end subscriptbe the distance between them. Then

    Potential difference between two equipotential surfaces A and B.
    V subscript A end subscript minus V subscript B end subscript equals k q open parentheses fraction numerator 1 over denominator r subscript A end subscript end fraction minus fraction numerator 1 over denominator r subscript B end subscript end fraction close parentheses
    equals k q open parentheses fraction numerator r subscript B end subscript minus r subscript A end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction close parentheses
    equals fraction numerator k q t subscript 1 end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction
    Or
    t subscript 1 end subscript equals fraction numerator open parentheses V subscript A end subscript minus V subscript B end subscript close parentheses r subscript A end subscript r subscript B end subscript over denominator k q end fraction
    Or t subscript 1 end subscript proportional to r subscript A end subscript r subscript B end subscript
    Similarly, t subscript 2 end subscript proportional to r subscript B end subscript r subscript C end subscript
    Since,
    r subscript A end subscript less than r subscript B end subscript less than r subscript C end subscript comma therefore r subscript A end subscript r subscript B end subscript less than r subscript B end subscript r subscript C end subscript
    therefore blank t subscript 1 end subscript less than t subscript 2 end subscript

    Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential differenceV subscript A end subscript minus V subscript B end subscript equals V subscript B end subscript minus V subscript C end subscript. If t subscript 1 end subscriptand t subscript 2 end subscriptbe the distance between them. Then

    physics-General
    Potential difference between two equipotential surfaces A and B.
    V subscript A end subscript minus V subscript B end subscript equals k q open parentheses fraction numerator 1 over denominator r subscript A end subscript end fraction minus fraction numerator 1 over denominator r subscript B end subscript end fraction close parentheses
    equals k q open parentheses fraction numerator r subscript B end subscript minus r subscript A end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction close parentheses
    equals fraction numerator k q t subscript 1 end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction
    Or
    t subscript 1 end subscript equals fraction numerator open parentheses V subscript A end subscript minus V subscript B end subscript close parentheses r subscript A end subscript r subscript B end subscript over denominator k q end fraction
    Or t subscript 1 end subscript proportional to r subscript A end subscript r subscript B end subscript
    Similarly, t subscript 2 end subscript proportional to r subscript B end subscript r subscript C end subscript
    Since,
    r subscript A end subscript less than r subscript B end subscript less than r subscript C end subscript comma therefore r subscript A end subscript r subscript B end subscript less than r subscript B end subscript r subscript C end subscript
    therefore blank t subscript 1 end subscript less than t subscript 2 end subscript
    General
    physics-

    Charges +q and –q are placed at points A and B respectively which are a distance 2 Lapart, C is the mid-point between A and B. The work done in moving a charge+Q along the semicircle CRD is

    In Ist case, when charge plus Q is situated at C

    Electric potential energy of system
    U subscript 1 end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses q close parentheses open parentheses negative q close parentheses over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses negative q close parentheses Q over denominator L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator L end fraction
    In IInd case, when charge plus Qis moved from C to D.

    Electric potential energy of system in that case
    U subscript 2 end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator open parentheses q close parentheses open parentheses negative q close parentheses over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator 3 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses negative q close parentheses open parentheses Q close parentheses over denominator L end fraction
    therefore W o r k blank d o n e equals increment U equals U subscript 2 end subscript minus U subscript 1 end subscript
    equals open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator 3 L end fraction minus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction close parentheses
    negative open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction close parentheses
    equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction. open square brackets fraction numerator 1 over denominator 3 L end fraction minus fraction numerator 1 over denominator L end fraction close square brackets
    equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses 1 minus 3 close parentheses over denominator 3 L end fraction
    equals fraction numerator negative 2 q Q over denominator 12 pi epsilon subscript 0 end subscript L end fraction equals negative fraction numerator q Q over denominator 6 pi epsilon subscript 0 end subscript L end fraction

    Charges +q and –q are placed at points A and B respectively which are a distance 2 Lapart, C is the mid-point between A and B. The work done in moving a charge+Q along the semicircle CRD is

    physics-General
    In Ist case, when charge plus Q is situated at C

    Electric potential energy of system
    U subscript 1 end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses q close parentheses open parentheses negative q close parentheses over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses negative q close parentheses Q over denominator L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator L end fraction
    In IInd case, when charge plus Qis moved from C to D.

    Electric potential energy of system in that case
    U subscript 2 end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator open parentheses q close parentheses open parentheses negative q close parentheses over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator 3 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses negative q close parentheses open parentheses Q close parentheses over denominator L end fraction
    therefore W o r k blank d o n e equals increment U equals U subscript 2 end subscript minus U subscript 1 end subscript
    equals open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator 3 L end fraction minus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction close parentheses
    negative open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction close parentheses
    equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction. open square brackets fraction numerator 1 over denominator 3 L end fraction minus fraction numerator 1 over denominator L end fraction close square brackets
    equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses 1 minus 3 close parentheses over denominator 3 L end fraction
    equals fraction numerator negative 2 q Q over denominator 12 pi epsilon subscript 0 end subscript L end fraction equals negative fraction numerator q Q over denominator 6 pi epsilon subscript 0 end subscript L end fraction
    parallel
    General
    physics-

    Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure left parenthesis a less than b less than c right parenthesis. Their surface charge densities are sigma comma negative sigma a n d sigma respectively. Calculate the electric potential on the surface of shell A

    The electric potential on the surface of shell
    V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript a end subscript over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction plus fraction numerator q subscript b end subscript over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript c end subscript over denominator c end fraction
    Or V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi a to the power of 2 end exponent sigma over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi b to the power of 2 end exponent open parentheses negative sigma close parentheses over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi c to the power of 2 end exponent sigma over denominator c end fraction
    open parentheses because q equals 4 pi r to the power of 2 end exponent sigma close parentheses
    o r blank V subscript A end subscript equals fraction numerator sigma over denominator epsilon subscript 0 end subscript end fraction left parenthesis a minus b plus c right parenthesis

    Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure left parenthesis a less than b less than c right parenthesis. Their surface charge densities are sigma comma negative sigma a n d sigma respectively. Calculate the electric potential on the surface of shell A