Maths-
General
Easy

Question

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

  1. 4,2
  2. 1,2
  3. 1,4
  4. 2,4

Hint:

In this question we have to find the order and degree of the given differential equation. For this we will first arrange the equation in a proper way. Now we know that the order of a differential equation is the order of the highest order derivative involved in the differential equation and the degree of a differential equation is the exponent of the highest order derivative involved in the differential equation.

The correct answer is: 2,4


    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

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    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

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    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
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    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
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    The charge deposited on 4 mu F capacitor the circuit is


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    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
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    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
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    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

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    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

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    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
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    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
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    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
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    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
    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

    physics-General
    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
    General
    physics-

    Work required to set up the four charge configuration (as shown in the figure) is

    Work is required to set up the four charge configuration
    q subscript 1 end subscript equals plus q comma blank q subscript 2 end subscript equals negative q comma blank q subscript 3 end subscript equals plus q a n d q subscript 4 end subscript equals negative q

    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator A B end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator B C end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator C D end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator D A end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses plus q close parentheses over denominator A C end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses negative q close parentheses over denominator B D end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets negative fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction close square brackets
    W equals 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 a end fraction open square brackets negative 4 plus square root of 2 close square brackets equals 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 a end fraction open square brackets negative 4 plus 1.414 close square brackets
    W equals negative 0.21 cross times fraction numerator q to the power of 2 end exponent over denominator epsilon subscript 0 end subscript a end fraction open parentheses a p p r o x. close parentheses

    Work required to set up the four charge configuration (as shown in the figure) is

    physics-General
    Work is required to set up the four charge configuration
    q subscript 1 end subscript equals plus q comma blank q subscript 2 end subscript equals negative q comma blank q subscript 3 end subscript equals plus q a n d q subscript 4 end subscript equals negative q

    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator A B end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator B C end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator C D end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator D A end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses plus q close parentheses over denominator A C end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses negative q close parentheses over denominator B D end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets negative fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction close square brackets
    W equals 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 a end fraction open square brackets negative 4 plus square root of 2 close square brackets equals 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 a end fraction open square brackets negative 4 plus 1.414 close square brackets
    W equals negative 0.21 cross times fraction numerator q to the power of 2 end exponent over denominator epsilon subscript 0 end subscript a end fraction open parentheses a p p r o x. close parentheses
    General
    physics-

    The points resembling equal potentials are

    The points S and R are inside the uniform electric field, so these will be at equal potential.

    The points resembling equal potentials are

    physics-General
    The points S and R are inside the uniform electric field, so these will be at equal potential.
    General
    physics-

    In the following diagram the work done in moving a point charge from point P to point A, B and C is respectively as W subscript A end subscript comma blank W subscript B end subscript a n d W subscript C end subscript then

    When a positive charge is moved from one point to another in an electric of magnetic field, then under the influence of the field force acts on the particle and an external agent will have to do work against this force. But in the given case the charge moves under influence of no field, hence it does not experience any force therefore, no work is done.
    W subscript A end subscript equals W subscript B end subscript equals W subscript C end subscript equals 0

    In the following diagram the work done in moving a point charge from point P to point A, B and C is respectively as W subscript A end subscript comma blank W subscript B end subscript a n d W subscript C end subscript then

    physics-General
    When a positive charge is moved from one point to another in an electric of magnetic field, then under the influence of the field force acts on the particle and an external agent will have to do work against this force. But in the given case the charge moves under influence of no field, hence it does not experience any force therefore, no work is done.
    W subscript A end subscript equals W subscript B end subscript equals W subscript C end subscript equals 0
    General
    physics-

    Three charges Q subscript 0 end subscript comma negative q and – q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q subscript 0 end subscriptis equal to

    Here total electrostatic potential energy is zero
    i e comma blank fraction numerator negative Q subscript 0 end subscript q over denominator l end fraction minus fraction numerator q Q subscript 0 end subscript over denominator l end fraction plus fraction numerator q to the power of 2 end exponent over denominator square root of 2 l end root end fraction equals 0
    On solving,
    Q subscript 0 end subscript equals fraction numerator q over denominator 2 square root of 2 end fraction equals fraction numerator 2 q over denominator square root of 32 end fraction

    Three charges Q subscript 0 end subscript comma negative q and – q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q subscript 0 end subscriptis equal to

    physics-General
    Here total electrostatic potential energy is zero
    i e comma blank fraction numerator negative Q subscript 0 end subscript q over denominator l end fraction minus fraction numerator q Q subscript 0 end subscript over denominator l end fraction plus fraction numerator q to the power of 2 end exponent over denominator square root of 2 l end root end fraction equals 0
    On solving,
    Q subscript 0 end subscript equals fraction numerator q over denominator 2 square root of 2 end fraction equals fraction numerator 2 q over denominator square root of 32 end fraction