Question

A spherical conductor has a radius of 14.0 cm and a charge of 42.0 µC. Calculate the electric field and the electric potential at the following distances from the center. (a) r = 8.0 cm magnitude direction electric field MN/C electric potential MV (b) r = 30.0 cm magnitude direction electric field MN/C electric potential MV (c) r = 14.0 cm magnitude direction electric field MN/C electric potential MV

Answer 1

Answer:

0 MN/C, 2.697 MV

4.1953 MN/C, 1.2586 MV

19.2642857143 MN/C, 2.697 MV

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field at r = 8 cm

E = 0 (inside)

Electric potential is given by

$V=\dfrac{kq}{R}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV$

Electric potential is 2.697 MV

r = 30 cm

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3^2}\\\Rightarrow E=4195333.33\ MN/C=4.1953\ MN/C$

Electric field is 4.1953 MN/C

$V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3}\\\Rightarrow V=1258600\ V=1.2586\ MV$

Electric potential is 1.2586 MV

r = R = 14 cm

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14^2}\\\Rightarrow E=19264285.7143\ N/C=19.2642857143\ MN/C$

The electric field is 19.2642857143 MN/C

$V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV$

The potential is 2.697 MV