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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 :

Answer:

Part a)

[tex]V = -1.52 V[/tex]

Part b)

[tex]V = -1.16 V[/tex]

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

[tex]V = \frac{kQ}{R}[/tex]

here we know that

[tex]Q = Q_1 - 6Q_1[/tex]

[tex]Q = - 5 Q_1[/tex]

[tex]Q_1 = 2.70 pC[/tex]

now we have

[tex]V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}[/tex]

[tex]V = -1.52 V[/tex]

Part b)

Potential on the axis of the ring is given as

[tex]V = \frac{kQ}{\sqrt{r^2 + R^2}}[/tex]

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

[tex]V = -1.16 V[/tex]

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