Chapter 22, Problem 75P

a.

To determine

The direction of the electric field at the center.

b.

To determine

Theelectric field at the center of the loop.

The electric field at the center $E=\frac{kQl}{2\pi a^3}$

Given:

  

The charges are placed as shown in the figure in the circular loop of radius a. Short gap of length l. the charge Q is evenly distributed on the non-conducting plastic rod.

Formula Used:

Electric field

  $E=\frac{kQ}{r^2}$

E is the electric field.

k is a constant.

Q is the charge.

r is the distance.

Calculations:

  $E=\frac{kQ}{r^2}$ (1)

The electric field at the center is only because of the loop radially opposite of the gap. The electric field by the rest of the loop is zero as it’s cancelled out by the radially opposite end.

Hence the electric field at the center is by the charge in the loop of length l radially opposite of the gap.

The charge at the at the loop of length l is

  $\begin{array}{l}q=\lambda l\\ \lambda =\frac{Q}{2\pi a}\\ q=\frac{Ql}{2\pi a}\end{array}$ linear charge density

Substituting the value of the charge in equation (1)

  $E=\frac{kQl}{2\pi a^3}$

Conclusion:

The electric field at the center $E=\frac{kQl}{2\pi a^3}$