Question

A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface of the rod (not near either end) is 22 kN/C , what is the volume charge density?

Answer 1

Answer:

$\rho = 7.35\times \muC/m^3$

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

  $E = \dfrac{Rl}{2\epsilon_0}$

 ε₀ is the permitivity of free space

  R is the radius of the rod

 and also,

$\rho=\dfrac{2E\epsilon_0}{R}$

ρ is the volume charge density

$\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}$

$\rho = 7.35\times 10^{-6}\ C/m^3$

$\rho = 7.35\times \muC/m^3$

Hence, the volume charge density is equal to $\rho = 7.35\times \muC/m^3$