- #1
Physicslearner500039
- 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.