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

Answer:

Part a)

$V = -1.52 V$

Part b)

$V = -1.16 V$

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

$V = \frac{kQ}{R}$

here we know that

$Q = Q_1 - 6Q_1$

$Q = - 5 Q_1$

$Q_1 = 2.70 pC$

now we have

$V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}$

$V = -1.52 V$

Part b)

Potential on the axis of the ring is given as

$V = \frac{kQ}{\sqrt{r^2 + R^2}}$

$V = \frac{(9\times 10^9)(-5\times 2.70\times 10^{-12})}{\sqrt{0.08^2 + 0.0671^2}}$

$V = -1.16 V$