Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring. (a) 18 pi (b) 24 pi (c) 36 pi (d) 72 pi?

Question

Answer

Finding Potential due to a Charged Ring

Given parameters:
- Density of the charged ring = 2 units
- Radius of the charged ring = 2m
- Distance of point from the ring = 1m

We need to find the potential due to the charged ring at the given point.

Formula for Potential due to a Charged Ring

The formula for potential due to a charged ring is given by:

V = k λ R^2 / (2R^2 + r^2 - 2Rr cosθ)^1/2

Where,
- V = potential
- k = Coulomb constant = 1/4πε
- λ = linear charge density
- R = radius of the ring
- r = distance of point from the center of the ring
- θ = angle between the axis of the ring and the line joining the center of the ring to the point

Solution

We are given that the density of the charged ring is 2 units. We can assume that this is the linear charge density (λ) of the ring.

The radius of the charged ring is 2m and the distance of the point from the ring is 1m. We can use these values to calculate the value of r.

r = 1m

Now, we need to find the value of θ. We can use trigonometry to calculate this value.

cosθ = R / √(R^2 + r^2)

cosθ = 2 / √(2^2 + 1^2)

cosθ = 2 / √5

θ = cos^-1(2 / √5)

θ = 1.107 radians

Now, we can substitute all the values in the formula for potential due to a charged ring.

V = k λ R^2 / (2R^2 + r^2 - 2Rr cosθ)^1/2

V = (1/4πε) (2) (2^2) / (2(2^2) + 1^2 - 2(2)(1) cos(1.107))^1/2

V = (2πε)^-1 (8 / 3)^1/2

V = 18π / (3√3)

V = 6√3π

Therefore, the potential due to the charged ring at the given point is 6√3π. The answer closest to this value is option (a) 18π.

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