Question

In Fig. 22-44, a thin glass rod forms a semicircle of radius r = 3.87 cm. Charge is uniformly distributed along the rod, with +q = 3.52 pC in the upper half and -q = -3.52 pC in the lower half. What is the magnitude of the electric field at P, the center of the semicircle?

Answer 1

Concepts and reason

The main concept used is to solve this problem is electric field at the center of a circular arc.

First calculate the electric field due to each quarter is found at the center.

Later, take the vector sum of electric field due to two quarter to calculate the net electric field.

Finally integrate the electric field expression over the full arc and plug in the values to calculate the magnitude of electric field.

Fundamentals

The electric field at the center of curvature due to charge distribution on a circular arc is given by following expression:

Here, is the permittivity of the free space, is the magnitude of charge, is the distance of the point from the charge, and is the angle.

Net electric field at the center is given by vector sum of electric field due to two quarters as follows:

$E=E,+E $

Here, and are the electric field at the center.

Find the electric field at the center of circular arc due to upper half.

Consider a small line element of angular width making angle with positive y- axis with uniform linear charge density .

Let q be the charge at the center of circular arc of radius r.

The electric field due to small line element is,

Linear charge density is given as follows:

So, the electric field due to small element is,

The horizontal component of electric field points in positive x-direction and the vertical component points in negative y- direction.

In vector form, electric field is given below:

As upper and lower half of the arc are symmetric with respect to horizontal, having same charge but of opposite sign, the electric field due to lower arc is given below:

Net electric field is given by the vector sum of the two fields as follows:

Find the electric field at the center of the semicircle.

Net electric field at the center due to small elements is,

Total electric field due to whole arc is,

Substitute for , for q , for , for and for r as follows:

Remove the suffices of the units as follows:

The electric field acts in negative y- direction and the magnitude of the electric field is .

Ans:

The magnitude of the electric field at the center of semi-circle is .