MCQMediumJEE 2026Electric Potential & Potential Energy

JEE Physics 2026 Question with Solution

There are three co-centric conducting spherical shells AA, BB and CC of radii aa, bb and cc respectively (c>b>a)(c>b>a) and they are charged with charges q1q_1, q2q_2 and q3q_3 respectively. The potentials of the spheres AA, BB and CC respectively are:

  • A

    14πε0(q1+q2+q3a),  14πε0(q1+q2+q3b),  14πε0(q1+q2+q3c)\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2+q_3}{a}\right),\; \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2+q_3}{b}\right),\; \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2+q_3}{c}\right)

  • B

    14πε0(q1a+q2b+q3c),  14πε0(q1+q2+q3b),  14πε0(q1+q2+q3c)\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{a}+\dfrac{q_2}{b}+\dfrac{q_3}{c}\right),\; \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2+q_3}{b}\right),\; \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2+q_3}{c}\right)

  • C

    14πε0(q1a+q2b+q3c),  14πε0(q1+q2b+q3c),  14πε0(q1+q2+q3c)\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{a}+\dfrac{q_2}{b}+\dfrac{q_3}{c}\right),\; \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2}{b}+\dfrac{q_3}{c}\right),\; \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2+q_3}{c}\right)

  • D

    14πε0(q1+q2+q3a),  14πε0(q1+q2b+q3c),  14πε0(q1a+q2b+q3c)\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2+q_3}{a}\right),\; \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2}{b}+\dfrac{q_3}{c}\right),\; \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{a}+\dfrac{q_2}{b}+\dfrac{q_3}{c}\right)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Three concentric conducting spherical shells AA, BB and CC have radii aa, bb and cc with c>b>ac>b>a, and charges q1q_1, q2q_2 and q3q_3 respectively.

Find: The potentials of shells AA, BB and CC.

For a charged conducting spherical shell, the potential outside is the same as that of a point charge at the center, and the potential at every point inside the shell is constant and equal to the surface potential.

So, at shell AA, we add the contribution of its own charge and the constant potentials due to the outer shells:

VA=14πε0(q1a+q2b+q3c)V_A = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{a} + \dfrac{q_2}{b} + \dfrac{q_3}{c}\right)

Stepwise Evaluation of Potentials

At shell BB, the charges on shells AA and BB contribute as if located at the center up to radius bb, while shell CC contributes a constant potential inside it:

VB=14πε0(q1+q2b+q3c)V_B = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2}{b} + \dfrac{q_3}{c}\right)

At the outermost shell CC, all charges act as if concentrated at the center:

VC=14πε0(q1+q2+q3c)V_C = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1+q_2+q_3}{c}\right)

Therefore, the correct expressions match Option (3). The correct option is C.

Common mistakes

  • A common mistake is to take the potential on every shell as depending on the total charge divided by that shell radius. This is wrong because outer shell charges give constant potential inside, not qr\dfrac{q}{r} with the inner radius. Use the surface radius of the shell producing the potential.

  • Another mistake is to ignore the contribution of outer shells to the potential at inner shells. Although the electric field inside a shell is zero, the potential is not zero there; it remains a constant equal to the shell surface potential.

  • Students may confuse electric field with electric potential. Zero electric field inside a conductor does not mean zero potential. Compute potential by adding scalar contributions from all shells.

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