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NVAMediumJEE 2023Electric Potential & Potential Energy

JEE Physics 2023 Question with Solution

Three concentric spherical metallic shells X, Y, and Z of radii aa, bb, and cc respectively (a<b<ca < b < c) have surface charge densities σ\sigma, −σ-\sigma, and σ\sigma respectively. The shells X and Z are at the same potential. If the radii of X and Y are 2 cm2 \, \text{cm} and 3 cm3 \, \text{cm} respectively, the radius of shell Z is:

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: Three concentric spherical metallic shells X, Y, and Z have radii aa, bb, and cc with surface charge densities σ\sigma, −σ-\sigma, and σ\sigma respectively. The shells X and Z are at the same potential. Also, a=2 cma = 2 \, \text{cm} and b=3 cmb = 3 \, \text{cm}.

Find: The value of cc.

For concentric spherical shells, the potential at a shell is obtained by superposition of contributions due to all shells.

The charges on the shells are:

qx=σ4πa2,qy=−σ4πb2,qz=σ4πc2q_x = \sigma 4\pi a^2, \qquad q_y = -\sigma 4\pi b^2, \qquad q_z = \sigma 4\pi c^2

The potential at shell Y is written as:

Vy=qx4πε0a+qy4πε0b+qz4πε0cV_y = \frac{q_x}{4\pi\varepsilon_0 a} + \frac{q_y}{4\pi\varepsilon_0 b} + \frac{q_z}{4\pi\varepsilon_0 c}

Substituting the charges:

Vy=σ4πa24πε0a−σ4πb24πε0b+σ4πc24πε0cV_y = \frac{\sigma 4\pi a^2}{4\pi\varepsilon_0 a} - \frac{\sigma 4\pi b^2}{4\pi\varepsilon_0 b} + \frac{\sigma 4\pi c^2}{4\pi\varepsilon_0 c}

So,

Vy=σaε0−σbε0+σcε0V_y = \frac{\sigma a}{\varepsilon_0} - \frac{\sigma b}{\varepsilon_0} + \frac{\sigma c}{\varepsilon_0}

Hence,

Vy=σε0(a−b+c)V_y = \frac{\sigma}{\varepsilon_0}(a - b + c)

Using the condition from the equality of potentials of shells X and Z:

c(a−b+c)=a2−b2+c2c(a - b + c) = a^2 - b^2 + c^2

Expanding,

c(a−b)+c2=a2−b2+c2c(a - b) + c^2 = a^2 - b^2 + c^2

Cancelling c2c^2 from both sides:

c(a−b)=a2−b2c(a - b) = a^2 - b^2

Factorizing the right-hand side:

c(a−b)=(a+b)(a−b)c(a - b) = (a + b)(a - b)

Therefore,

c=a+bc = a + b

Substituting a=2 cma = 2 \, \text{cm} and b=3 cmb = 3 \, \text{cm}:

c=2+3=5 cmc = 2 + 3 = 5 \, \text{cm}

Therefore, the radius of shell Z is 5 cm5 \, \text{cm}.

Common mistakes

  • Using electric field formulas instead of potential superposition. That is incorrect because the question compares potentials of shells, not fields at a point. Write the potential due to each spherical shell and then add the contributions.

  • Assigning the charge on shell Y as positive. This is wrong because its surface charge density is −σ-\sigma, so its contribution to potential carries a negative sign. Keep the sign of each shell charge throughout the algebra.

  • Using the wrong potential expression for a spherical shell by treating every contribution as if the observation point were outside all shells. For a shell, the potential at any point inside it is constant and equals the surface potential. Use the appropriate radius in each term exactly as written in the solution.

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