A charge of +3.0 μC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energy is required to bring a charge of 25μC from infinity to the center of the circle?
A) 5.4 J
B) 3.4 J
C) 4.3 J
D) 2.7 J
E) 6.8 J

Question

Quizplus · Accepted Answer

The electric potential at the center of the circle due to the distribution of charge on the circumference is given by:
$$V=\frac{kQ}{r}$$
where k is the Coulomb constant, Q is the total charge distributed along the circumference and r is the radius of the circle.
We have Q = +3.0 μC and r = 20 cm = 0.2 m.
Therefore, 
$$V=\frac{(9	imes10^9\ Nm^2/C^2)(3.0	imes10^{-6}C)}{0.2\ m}=1.35	imes10^5\ V$$
Now, to bring a charge of 25μC from infinity to the center of the circle, we need to do work against the electric field. The work done in moving a charge q through a potential difference V is given by:
$$W=qV$$
Plugging in the values, we get:
$$W=(25	imes10^{-6}C)(1.35	imes10^5\ V)=3.38\ J$$
Therefore, the external energy required is 3.4 J (to two significant figures).

A Charge of +3