Question

The dome of a Van de Graaff generator receives a charge of 7.0 x 10^−4 C. Find the strength of the electric field in the following situations.
(Hint: Review properties of conductors in electrostatic equilibrium. Also, use the points on the surface are outside a spherically symmetric charge distribution; the total charge may be considered to be located at the center of the sphere.)
(a) inside the dome in N/C(b) at the surface of the dome, assuming it has a radius of 3.0 m in N/C
(c)8.0 m from the center of thedome in N/C

Answer 1

Answer:

0

77691.3580247 N/C

98328.125 N/C

Explanation:

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

q = Charge = $7\times 10^{-4}\ C$

r = Distance

In a Van de Graff generator there is no charge inside the dome.

So, the electric field inside the dome is 0

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 7\times 10^{-4}}{9^2}\\\Rightarrow E=77691.3580247\ N/C$

The electric field at the surface is 77691.3580247 N/C

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 7\times 10^{-4}}{8^2}\\\Rightarrow E=98328.125\ N/C$

The electric field at the center of the dome is 98328.125 N/C