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NVAMediumJEE 2023Gauss's Law Applications

JEE Physics 2023 Question with Solution

For a charged spherical ball, the electrostatic potential inside the ball varies with rr as V=2ar2+bV = 2ar^2 + b. The volume charge density inside the ball is −λaε0-\lambda a \varepsilon_0. The value of λ\lambda is:

Answer

Correct answer:12

Step-by-step solution

Standard Method

Given: The potential inside the spherical ball is V=2ar2+bV = 2ar^2 + b.

Find: The value of λ\lambda if the volume charge density is ρ=−λaε0\rho = -\lambda a \varepsilon_0.

The electric field is related to potential by

E=−dVdr=−ddr(2ar2+b)=−4arE = -\frac{dV}{dr} = -\frac{d}{dr}(2ar^2 + b) = -4ar

For spherical symmetry, the charge density is

ρ=ε0∇⋅E=ε0⋅1r2ddr(r2E)\rho = \varepsilon_0 \nabla \cdot \mathbf{E} = \varepsilon_0 \cdot \frac{1}{r^2} \frac{d}{dr}(r^2 E)

Substitute E=−4arE = -4ar:

ρ=ε0⋅1r2ddr(−4ar3)\rho = \varepsilon_0 \cdot \frac{1}{r^2} \frac{d}{dr}(-4ar^3) ρ=ε0⋅1r2(−12ar2)=−12aε0\rho = \varepsilon_0 \cdot \frac{1}{r^2}(-12ar^2) = -12a\varepsilon_0

Comparing with ρ=−λaε0\rho = -\lambda a \varepsilon_0, we get

λ=12\lambda = 12

Therefore, the value of λ\lambda is 1212.

Common mistakes

  • Using E=dVdrE = \frac{dV}{dr} instead of E=−dVdrE = -\frac{dV}{dr}. The negative sign is essential because electric field points in the direction of decreasing potential. Always differentiate with the minus sign first.

  • Using the Cartesian form of divergence directly for a spherically symmetric field. Here EE depends only on rr, so the correct form is ∇⋅E=1r2ddr(r2E)\nabla \cdot \mathbf{E} = \frac{1}{r^2}\frac{d}{dr}(r^2E).

  • Comparing ρ=−12aε0\rho = -12a\varepsilon_0 incorrectly with −λaε0-\lambda a \varepsilon_0 and missing that the factor of aε0a\varepsilon_0 is common on both sides. Cancel the common factors carefully to obtain λ=12\lambda = 12.

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