Question

A total charge of $5\mu c$ is distributed uniformally on the surface of the thin walled semi spherical cup. If the electric field strength at the center of the semi sphere is $9\times {10}^{8}N{C}^{-1}$, the radius of the cup is:
$(\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N{m}^2{C}^{-2})$

• A. 5 mm
• B. 10 mm
• C. 5 cm
• D. 10 cm

Answer 1

The correct option is A 5 mm

Electric field at the center of a semispherical cup of surface charge density $\sigma =\frac{\sigma }{4{ϵ}_0}=9\times {10}^8$

$⟹\sigma =4{ϵ}_0\times 9\times {10}^8=\frac{1}{10\pi }$

$⟹\frac{q}{2\pi R^2}=\frac{1}{10\pi }$

$⟹R=\sqrt{5\times 5\times {10}^{-6}}=5mm$