Question

A uniformly charged insulating rod of length 10.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of −7.50 µC.

(a) Find the magnitude of the electric field at O, the center of the semicircle.   

      =           N/C. 

 

Answer 1

Concepts and reason

The concept used to solve this problem is electric field at the centre of the semicircle.

Initially find out the horizontal component of electric field and use the relation between linear charge density and radius of the semicircle for charge in the expression.

To find the net field, integrate the electric field expression from 0 to , substitute the values in the expression and find the value of electric field.

Fundamentals

Expression for the charge is,

Here, is the charge, is the linear charge density, is the radius of the semicircle, and is the angle.

Expression for the radius of the semicircle is,

Here, L is the length of the rod.

Expression for the linear charge density is,

Expression for the horizontal component of electric field is,

Substitute for .

Integrate the above equation with to 0 limits.

On integration we get,

Substitute for r and $7/b=r $for .

Expression for the magnitude of the electric field at O, the center of the semicircle is,

Substitute for , for q, and for L.

Ans:

The magnitude of the electric field at O, the centre of the semicircle is.