Answer to Question #162352 in Electric Circuits for Polycarp bogonko

Question

a)an isolated conducting sphere of 10cm placed in a vacuum carried a positive charge of 1.0×10^-10coulombs calculate it's electrical intensity.

b)the p.d in volts at the surface of the sphere and 5m from the sphere respectfully.

Answer 1

(a) radius of sphere(R) =0.1m

at the surface of conducting sphere

Electric field intensity can be given as

$E=\frac{kq}{R^2}$

K is electrostatic constant which value for vaccum is $9\times10^9 Nm^2/C^2$

By putting value

$E=\frac{(9\times10^9)(1.0\times10^{-10})}{(0.1)^2}\\E=90N/C$

(B) potential difference can be given as

$\Delta V=kq(\frac{1}{R}-\frac{1}{r})$

$\Delta V=(9\times10^{9})(10^{-10})(\frac{1}{0.1}-\frac{1}{5})$

$\Delta V=8.82volt$