Question

A plastic rod has been bent into a circle of radius R = 8.00 cm. It has a charge Q1 = +2.70 pC uniformly distributed along one-quarter of its circumference and a charge Q2 = −6Q1 uniformly distributed along the rest of the circumference. Take V = 0 at infinity. (a) What is the electric potential at the center C of the circle? V (b) What is the electric potential at point P, which is on the central axis of the circle at distance D = 6.71 cm from the center?

Answer 1

Part a)

`V = -1.52 V`

Part b)

`V = -1.16 V`

Step-by-step explanation:

Part a)

Electric potential is a scalar quantity

so here we can say that total potential due to a ring on its center is given as

`V = (kQ)/(R)`

here we know that

`Q = Q_1 - 6Q_1`

`Q = - 5 Q_1`

`Q_1 = 2.70 pC`

now we have

`V = ((9* 10^9)(-5* 2.70 * 10^(-12)))/(0.08)`

`V = -1.52 V`

Part b)

Potential on the axis of the ring is given as

`V = (kQ)/(√(r^2 + R^2))`

`V = ((9* 10^9)(-5* 2.70* 10^(-12)))/(√(0.08^2 + 0.0671^2))`

`V = -1.16 V`