Question

A sphere of radius r₁ = 30 cm has a charge density \(\left(\rho\dfrac{r}{r_1}\right)\) where ρ₀ = 200 pC/m³. The value of total change on the sphere is given by
1. 34 pC
2. 8.5 pC
3. 17 pC
4. 68 pC

Answer 1

Correct Answer - Option 3 : 17 pC

Concept:

Gauss law states that Aux leaving through any closed surface is equal to the charge enclosed by that surface i.e

\(\psi = \mathop \oint \nolimits_s^{} D.ds = Q = \mathop \smallint \nolimits_v^{} {\rho _v}\;dv\)

ψ = flux

ρ_v = volume charge density

D = Electric flux density

ρ_v = ∇ .D (Gauss law in differential form).

Gauss law is applicable for time-varying as well as the static field.

The equation is valid irrespective of the shape of closed surface areas.

Analysis:

Given radius r₁ = 30 cm, \({\rho _v} = \frac{{{\rho _r}}}{{{r_1}}}\), ρ₀ = 20 pC/m³

By using Gauss law i.e.

\(Q = \;\mathop \smallint \limits_v^{} {\rho _v}\;dv\)

\(Q = \int\int\int\;\frac{{{\rho _r}}}{{{r_1}}}.{r^2}\sin \theta \;dr\;d\theta \;d\phi \)

( 0 ≤ r ≤ r₁, 0 ≤ θ ≤ π, 0 ≤ ϕ ≤ 2π)

\(Q = \mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^\pi \mathop \smallint \limits_0^r \frac{{\rho r}}{{{r_1}}}{r^2}\sin d\theta \;dr\;d\theta \;d\phi \)

\(Q = \frac{{{\rho _0}}}{{{r_1}}}\mathop \smallint \limits_0^{2\pi } d\phi \mathop \smallint \limits_0^\pi \sin \theta \;d\theta \mathop \smallint \limits_0^{{r_1}} {r^3}dr\)

\(Q = \frac{{{\rho _0}}}{{{r_1}}}\left[ {2\pi } \right]\;\left[ { - \cos \theta } \right]_0^\pi \;\left[ {\frac{{{r^4}}}{4}} \right]_0^{{r_1}}\)

\(Q = \frac{{2\pi {\rho _0}}}{{{r_1}}}\left[ 2 \right]\left( {\frac{{r_1^4}}{4}} \right)\)

\(Q = \frac{{4\pi {\rho _0}r_1^3}}{4} = \pi {\rho _0}r_1^3\)

\(Q = \pi {\rho _0}r_1^3\)

Substitute values, we get:

Q = TT × 200 (PC/m³) × (30 × 10^-2)³

Q = 17 PC

Option (C) correct.