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Easy

Question

The order of differential equation  isopen parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed equals open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 1 divided by 2 end exponent

  1. 2
  2. 3
  3. 1 half
  4. 6

Hint:

In this question we have to find the order 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.

The correct answer is: 2


    open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed equals open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 1 divided by 2 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 6 equals 1 plus fraction numerator d y over denominator d x end fraction
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 6 minus fraction numerator d y over denominator d x end fraction equals 1
    order = 2

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    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
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    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
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    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

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    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

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    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

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    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
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    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
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    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
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    The charge deposited on 4 mu F capacitor the circuit is


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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
    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
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    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
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    Q equals blank 3 blank mu F cross times 12 V equals 36 mu C

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    Also Q equals Q subscript 1 end subscript plus Q subscript 2 end subscript
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    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

    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

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    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?

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    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

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    General
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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

    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
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    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
    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-