Question

shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; ring 2 has uniform charge q2 and the same radius R. The rings are separated by distance d = 3.00R. The net electric field at point P on the common line, at distance R from ring 1, is zero. What is the ratio q1/q2?

Answer 1

Answer:

$\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}$

Explanation:

Given that

Charge on ring 1 is q1 and radius is R.

Charge on ring 2 is q2 and radius is R.

Distance ,d= 3 R

So the total electric field at point P is given as follows

Given that distance from ring 1 is R

$\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(d-R)}{(R^2+(d-R)^2)^{3/2}}=0$

$\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(3R-R)}{(R^2+4R^2)^{3/2}}=0$

$\dfrac{q_1R}{(2R^2)^{3/2}}-\dfrac{q_2(2R)}{(5R^2)^{3/2}}=0$

$\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}$

Answer 2

The figure is missing, so i have attached it.

Answer:

q1/q2 = 2[(2/5)^(3/2)] ≈ 0.506

Explanation:

The Electric Field for a ring of charge is given by the equation;

E = qz/[4πε_o(z² + R²)]

where;

z is the distance along the z-axis

q is the charge on the ring

R is the radius of the ring.

Now, at the point P, we want the two contributions to be equal to each other.

So, plugging in the same R

for each and the appropriate z, we get;

q1•R/[4πε_o(R² + R²)^(3/2)]

= q2•2R/[4πε_o((2R)² + R²)^(3/2)]

The terms 4πε_o cancels out to give ;

q1•R/[(R² + R²)^(3/2)]

= q2•2R/[((4R² + R²)^(3/2)]

Let's rearrange to get q1/q2.

Thus;

q1/q2 = 2R/[((4R² + R²)^(3/2)]/[R/[(R² + R²)^(3/2)]]

This gives;

q1/q2 = 2R/[((4R² + R²)^(3/2)]•[[(R² + R²)^(3/2)]/R]

Gives;

q1/q2 = [2/[(2R²)^(3/2)]]/[(5R²)^(3/2)]

Gives;

q1/q2 = 2(2R²/5R²)^(3/2)

R² cancels out to give ;

q1/q2 = 2[(2/5)^(3/2)] ≈ 0.506