Field strength from rod

Question

+6.40 nC of charge is uniformly distributed along the left half of a thin rod of total length L = 4.00 m, while -6.40 nC of charge is uniformly distributed along the right half of the rod. What is the magnitude of the electric field E at the dot, a distance r = 1.45 m from the centre of the rod?

Answer 1

Let's set the midpoint of the rod as x=0. If we draw E vectors from -x and x at the point, we see that the vector components in y cancel and the components in x add. We can figure out the E_x from adding up the dE_x from the dq along the line for -2≤x≤0 and double that to find the contribution from both parts of line charge.

The E_x component wil be Ecosθ for the triangle from the -x position on the line to the point (vertical position 1.45 m) The cos of the angle will be x/sqrt(x² + 1.45²)

It is also convenient to express dq as λdx where λ = 6.4 nC/2 m = 3.2 nC/m = 3.2 x 10^-9 C/m

Recall that dE = (1/4πε₀)dq/r²

The integral becomes (1/4πε₀) integral from x = -2 to 0 of (λdx/(x²+1.45²)(x/(x²+1.45²))

This reduces to (3.2 x 10^-9)/4πε₀) integral from -2 to 0 of xdx/(x²+1.45²)² which is integrable by u sub.

Please consider a tutor. Take care (Remember to integral by 2 to obtain contribution from both parts of the line charge)