An insulating solid sphere of the radius R is charged in a non-uniform manner such that the volume charge density ρ =(A/r) , where A is a positive constant and r is the distance from the centre. The potential difference between the centre and surface of the sphere is

Question

An insulating solid sphere of the radius R is charged in a non-uniform manner such that the volume charge density ρ =(A/r) , where A is a positive constant and r is the distance from the centre. The potential difference between the centre and surface of the sphere is (A) (A R/8 ϵ 0) (B) (A R/4 ϵ 0) (C) (A R/ ϵ 0) (D) (A R/2 ϵ 0)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-pni0xsxrdfng.png" /> E × 4π r2=(q/ϵ 0) E× 4π r2= displaystyle ∫ 0r (A/r)× 4π r2dr E=(A/2 ϵ 0) Since electric field is constant, we get Δ V=E× d =(A/2 ϵ 0)× R=(A R/2 ϵ 0)

Q. An insulating solid sphere of the radius $R$ is charged in a non-uniform manner such that the volume charge density $ρ = \frac{A}{r}$ , where $A$ is a positive constant and $r$ is the distance from the centre. The potential difference between the centre and surface of the sphere is

$\frac{A R}{8 ϵ _{0}}$

$\frac{A R}{4 ϵ _{0}}$

$\frac{A R}{ϵ _{0}}$

$\frac{A R}{2 ϵ _{0}}$

$E \times 4 π r^{2} = \frac{q}{ϵ _{0}}$
$E \times 4 π r^{2} = \int_{0}^{r} \frac{A}{r} \times 4 π r^{2} d r$
$E = \frac{A}{2 ϵ _{0}}$
Since electric field is constant, we get
$Δ V = E \times d$
$= \frac{A}{2 ϵ _{0}} \times R = \frac{A R}{2 ϵ _{0}}$

Q. An insulating solid sphere of the radius R is charged in a non-uniform manner such that the volume charge density ρ=rA​ , where A is a positive constant and r is the distance from the centre. The potential difference between the centre and surface of the sphere is

8ϵ0​AR​

4ϵ0​AR​

ϵ0​AR​

2ϵ0​AR​