Poll Results
No votes. Be the first one to vote.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
d
Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r
= 1m. We get V = 72π volts
To find the electric potential (V) at a point due to a charged ring, we can use the formula for the potential due to a ring of charge. Given a ring with linear charge density (lambda) (charge per unit length) and radius (R), and a point along the axis perpendicular to the plane of the ring and passing through its center at a distance (x) from the center of the ring, the electric potential (V) at that point is given by:
[ V = frac{kQ}{sqrt{R^2 + x^2}} ]
where (k) is Coulomb’s constant ((k approx 8.987 times 10^9 , text{Nm}^2/text{C}^2)), (Q) is the total charge on the ring, (R) is the radius of the ring, and (x) is the distance of the point from the center of the ring along the axis. The total charge (Q) can be found from the charge density (lambda) and the circumference of the ring ((Q = lambda cdot 2pi R)).
Given that the linear charge density (lambda = 2) units and the radius of the ring (R = 2) meters, and the point is at a distance of (x = 1) meter from the center of the ring along its axis, let’s calculate the potential