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JEE Physics 2025 Question with Solution

Match List - I with List - II: List - I: (A) Electric field inside (distance r>0r > 0 from center) of a uniformly charged spherical shell with surface charge density σ\sigma, and radius RR. (B) Electric field at distance r>0r > 0 from a uniformly charged infinite plane sheet with surface charge density σ\sigma. (C) Electric field outside (distance r>0r > 0 from center) of a uniformly charged spherical shell with surface charge density σ\sigma, and radius RR. (D) Electric field between two oppositely charged infinite plane parallel sheets with uniform surface charge density σ\sigma. List - II: (I) σϵ0\frac{\sigma}{\epsilon_0} (II) σ2ϵ0\frac{\sigma}{2\epsilon_0} (III) 00 (IV) σϵ0r2\frac{\sigma}{\epsilon_0 r^2} Choose the correct answer from the options given below:

  • A

    (A)-(IV),(B)-(II),(C)-(III),(D)-(I)\text{(A)-(IV)}, \text{(B)-(II)}, \text{(C)-(III)}, \text{(D)-(I)}

  • B

    (A)-(IV),(B)-(I),(C)-(III),(D)-(II)\text{(A)-(IV)}, \text{(B)-(I)}, \text{(C)-(III)}, \text{(D)-(II)}

  • C

    (A)-(III),(B)-(II),(C)-(IV),(D)-(I)\text{(A)-(III)}, \text{(B)-(II)}, \text{(C)-(IV)}, \text{(D)-(I)}

  • D

    (A)-(I),(B)-(II),(C)-(IV),(D)-(III)\text{(A)-(I)}, \text{(B)-(II)}, \text{(C)-(IV)}, \text{(D)-(III)}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Four electrostatic configurations are to be matched with their electric field expressions.

Find: The correct matching between List - I and List - II.

Use Gauss's law for each case.

∮E⃗⋅dA⃗=qenclosedε0\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}

For (A), inside a uniformly charged spherical shell, the enclosed charge is zero. Therefore,

E=0E = 0

So (A) \rightarrow (III).

For (B), the electric field due to a uniformly charged infinite plane sheet is constant:

E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}

So (B) \rightarrow (II).

For (C), outside a uniformly charged spherical shell, the field is the same as that of a point charge at the center. Since

Q=4πR2σQ = 4\pi R^2 \sigma

we get

E=14πε0Qr2=14πε04πR2σr2E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2} = \frac{1}{4\pi \varepsilon_0} \frac{4\pi R^2 \sigma}{r^2}

Thus the correct outside field is actually

E=σR2ε0r2E = \frac{\sigma R^2}{\varepsilon_0 r^2}

However, the solution matches this case with (IV) as listed on the solution's. So (C) \rightarrow (IV) according to the provided solution and options.

For (D), between two oppositely charged infinite parallel sheets, the fields add:

E=σ2ε0+σ2ε0=σε0E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}

So (D) \rightarrow (I).

Hence the final matching is:

(A)-(III), (B)-(II), (C)-(IV), (D)-(I)\text{(A)-(III), (B)-(II), (C)-(IV), (D)-(I)}

Therefore, the correct option is C.

Case-by-case Matching

Given: A match-the-following question on electric fields due to spherical shells and infinite sheets.

Find: Which option gives the correct pairings.

  1. Spherical shell, inside region: By symmetry and Gauss's law, no charge is enclosed for a Gaussian surface inside the shell.
E=0E = 0

Hence (A) \rightarrow (III).

  1. Single infinite plane sheet: The electric field on either side of a uniformly charged infinite sheet is uniform.
E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}

Hence (B) \rightarrow (II).

  1. Spherical shell, outside region: The shell behaves like a point charge with total charge
Q=4πR2σQ = 4\pi R^2 \sigma

So,

E=14πε0Qr2E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2}

Substituting for QQ,

E=14πε04πR2σr2=σR2ε0r2E = \frac{1}{4\pi \varepsilon_0} \frac{4\pi R^2 \sigma}{r^2} = \frac{\sigma R^2}{\varepsilon_0 r^2}

The source solution still maps this to (IV), so the intended matching is (C) \rightarrow (IV).

  1. Two oppositely charged infinite parallel sheets: Between the sheets, the fields due to the two sheets are in the same direction and add.
E=σ2ε0+σ2ε0=σε0E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}

Hence (D) \rightarrow (I).

So the complete match is:

  • (A) - (III)
  • (B) - (II)
  • (C) - (IV)
  • (D) - (I)

Therefore, the correct option is C.

Common mistakes

  • Students often think the electric field inside a uniformly charged spherical shell is non-zero because the shell carries charge. This is wrong because a Gaussian surface inside the shell encloses zero charge. Use Gauss's law and symmetry to conclude that the field inside is 00.

  • A common mistake is taking the field of a single infinite plane sheet as σε0\frac{\sigma}{\varepsilon_0}. That value applies to the region between two oppositely charged sheets, not one sheet alone. For one sheet, use σ2ε0\frac{\sigma}{2\varepsilon_0}.

  • Students may fail to add the fields between two oppositely charged parallel sheets. The two fields are in the same direction in the region between the sheets, so they add to give σε0\frac{\sigma}{\varepsilon_0}, not cancel.

  • For the field outside the spherical shell, students may overlook that the shell acts like a point charge with total charge Q=4πR2σQ = 4\pi R^2\sigma. The physically correct expression becomes σR2ε0r2\frac{\sigma R^2}{\varepsilon_0 r^2}, so one should be alert to dimensional consistency when reading the listed expressions.

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