A circular plastic rod of radius R = 2.40 m has a positive charge = Q = 8.00×10-

Question

A circular plastic rod of radius R = 2.40 m has a positive charge = Q = 8.00×10-6 C uniformly distributed along one-quarter of its circumference and a negative charge of -8Q uniformly distributed along the rest of the circumference.(See the figure) (At infinity V is zero).

A) What is the electric potential at the center C of the circle?

B) What is the electric potential at point P, which is on the central axis of the circle at a distance z = 1.632 m from the center?

Any help with the integration involved in this problem would be greatly appreciated!

Explanation / Answer

Here,

R = 2.4 m

Q = 8 *10^-6 C

A)

for electric potential at the centre

electric potential at the centre = k * (-8 Q)/R + k *Q/R

electric potential at the centre = - 7 * k *Q/R

electric potential at the centre = -7 * 9*10^9 * 8*10^-6/2.4

electric potential at the centre = -210000 V

the electric potential at the centre is -210000 V

B)

for point P

distance from ring , d = sqrt(1.632^2 + 2.4^2)

d = 2.902 m

electric potential at point P = k * (-8Q)/d + k * Q/d

electric potential at point P = -7 * k *Q/d

electric potential at point P = -7 * 9*10^9 * 8*10^-6/2.902

electric potential at point P is - 1.736 *10^5 V

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there is no integration involved in this problem , as all the charge is placed at same distance from the point

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