A sphere of radius R has a volume density of charge ρ = kr, where r is the distance from the centre of the sphere and k is constant. The magnitude of the electric field which exists at the sur ace o the sphere is given by (ε0 = permittivity of the free space)

Question

A sphere of radius R has a volume density of charge ρ = kr, where r is the distance from the centre of the sphere and k is constant. The magnitude of the electric field which exists at the sur ace o the sphere is given by (ε0 = permittivity of the free space) (A) (4π kR4/3ε0) (B) (kR/3ε0) (C) (4π kR/ε0) (D) (kR2/4ε0)

Tardigrade Admin · Accepted Answer

Given, ρ=K ⋅ r By Gauss's theorem E(4 π r2) =(∫ ρ × 4 π r2 d r/ε0) =(∫ K r × 4 π r2 d r/ε0) ⇒ E =(K r2/4 ε0) Here r =R So, E=(K R2/4 ε0)

Q. A sphere of radius R has a volume density of charge $ρ = k r$ , where r is the distance from the centre of the sphere and k is constant. The magnitude of the electric field which exists at the sur ace o the sphere is given by ( $ε_{0}$ = permittivity of the free space)

$\frac{4 πk R ^{4}}{3 ε _{0}}$

$\frac{k R}{3 ε _{0}}$

$\frac{4 πk R}{ε _{0}}$

$\frac{k R ^{2}}{4 ε _{0}}$

Given, $ρ = K \cdot r$
By Gauss's theorem
$E (4 π r^{2}) = \frac{\int ρ \times 4 π r ^{2} d r}{ε _{0}}$
$= \frac{\int Kr \times 4 π r ^{2} d r}{ε _{0}}$
$\Rightarrow E = \frac{K r ^{2}}{4 ε _{0}}$
Here $r = R$
So, $E = \frac{K R ^{2}}{4 ε _{0}}$

Q. A sphere of radius R has a volume density of charge ρ=kr, where r is the distance from the centre of the sphere and k is constant. The magnitude of the electric field which exists at the sur ace o the sphere is given by (ε0​ = permittivity of the free space)

3ε0​4πkR4​

3ε0​kR​

ε0​4πkR​

4ε0​kR2​