Question

Positive charge is placed on two conducting spheres that are very far apart and connected by a long, very thin conducting wire. The radius of the smaller sphere is 5.60 cm and the radius of the larger sphere is 10.5 cm. The electric field strength at the surface of the larger sphere is 215 kV/m. Find the surface charge density on the 5.60-cm sphere. (in C/m^2)

Answer 1

Answer:

$3.56\mu C/m^2$

Explanation:

The potential at the surface of the spheres is same as they are connected with a conducting wire.

Calculate electric field strength at the surface of smaller sphere.

$R_s=5.60 cm\\R_L=10.5 cm\\E_L=215 kV/m\\ \frac{E_s}{E_L}=\frac{R_L}{R_s}\\E_s= \frac{10.5}{5.60}\times 215\times 10^3 =4.03\times10^5V/m$

$\sigma_s=\epsilon_oE_s\\\sigma_s=8.85\times10^{-12}\times 4.03\times10^5\\\sigma_s=35.66\times 10^{-7}C/m^2$