- #1

- 124

- 6

- Homework Statement:
- A point charge q is located at the center of a thin ring of radius R with uniformly distributed charge —q. Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance x from its center, if x » R.

- Relevant Equations:
- NA

I thought it was easy but i am not getting the correct answer

The electric field due the point charge q is

##

E1 = q/(4\pi\epsilon x^2)

##

The electric field due to the thin ring of radius R is considering the electric field due to the element charge dq (dS)

##

dE2 = dq/4\pi\epsilon (x^2 + R^2) \\

dE2 = \lambda * dS/(4\pi\epsilon (x^2 + R^2)) \\

dE2_{net} = dE2\cos{\theta} \\

dE2_{net} = \lambda * dS * \cos{\theta} /(4\pi\epsilon (x^2 + R^2)) \\

\cos\theta = x/(x^2 + R^2)^\frac 1 2 \\

\lambda * dS * x /(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) \\

\int_0^{2\pi R} \lambda dSx/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) \\

\lambda x/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2)\int_0^{2\pi R} dS\\

\lambda x 2\pi R/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) \\

E2_{net} = qx/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) \\

##

The net field is

##

E2_{net} - E1 = qx/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) - q/(4\pi\epsilon x^2)

##

For ## x > R ## the net field is 0.

The electric field due the point charge q is

##

E1 = q/(4\pi\epsilon x^2)

##

The electric field due to the thin ring of radius R is considering the electric field due to the element charge dq (dS)

##

dE2 = dq/4\pi\epsilon (x^2 + R^2) \\

dE2 = \lambda * dS/(4\pi\epsilon (x^2 + R^2)) \\

dE2_{net} = dE2\cos{\theta} \\

dE2_{net} = \lambda * dS * \cos{\theta} /(4\pi\epsilon (x^2 + R^2)) \\

\cos\theta = x/(x^2 + R^2)^\frac 1 2 \\

\lambda * dS * x /(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) \\

\int_0^{2\pi R} \lambda dSx/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) \\

\lambda x/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2)\int_0^{2\pi R} dS\\

\lambda x 2\pi R/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) \\

E2_{net} = qx/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) \\

##

The net field is

##

E2_{net} - E1 = qx/(4\pi\epsilon (x^2 + R^2)^ \frac 3 2) - q/(4\pi\epsilon x^2)

##

For ## x > R ## the net field is 0.