Question

A charge of −25µC is distributed uniformly over the surface of a spherical conductor of radius 12.0 cm. Determine the electric field (in N/C) due to this charge at the following distances from the center of the sphere. (Enter the radial component of the electric field.)a) 5cmb) 10cmc) 50cm

Answer 1

Answer:

a) 0 b) 0 c) -9*10⁵ N/C

Explanation:

a)

• In electrostatic conditions, no electric field can exist inside a conductor.
• As the distance r=5 cm falls inside the conductor, the electric field is just zero.

b)

• Same as above, as r=10 cm is still inside the spherical conductor.

c)

• At r= 50 cm. from the center of the spherical conductor, we can apply Gauss' Law in order to get the value of the electric field.
• By symmetry, the electric field, at a same distance from the center, must be radial, and constant on a spherical surface concentric with the spherical conductor.
• So, we can write the following equation for Gauss'Law:

       $\int\ {E} \, dA = \frac{Q}{\epsilon_{0} }$

• If E is constant, we can take it out of the integral, and integrate all the closed spherical surface, as follows:

       $E* 4*\pi *r^{2} =\frac{Q}{\epsilon_{0}}$

• So, we can solve for E, as follows:

       $E = \frac{Q}{4*\pi*r^{2}*\epsilon_{0}} = \frac{(-25e-6)C}{4*\pi*(0.5m)^{2}*8.85e-12C2/N*m2}}\\\\ E = -9e5 N/C$

• E = -9*10⁵ N/C (radially inward, taking the outward direction as positive)