Question

A positive charge Q is distributed uniformly along the positive y-axis between y=0 and y=a. A negative point charge, -q, lies on the positive x-axis, a distance x from the origin. Calculate the x and y components of the E-field produced by the charge distribution Q at points on the positive x axis.

Answer 1

This can be done using integration and vector analysis. By considering a differential element of the line charge distribution dq = Q/a*dy, we could calculate the dFx and dFy and solve for Fx and Fy through integration.

dFx = -xkQdq / (y^2 + x^2)^3/2 = -xkQ^2/a/(y^2+x^2)^(3/2)*dy
Fx = integral from 0 to a -xkQ^2/a/(y^2+x^2)^(3/2)*dy
Fx = -xkQ^2/a*int(0, a, dy/(y^2+x^2)^(3/2))
Fx = -kQ^2/x(a^2+x^2)^(1/2)

dFy = kQy/(y^2+x^2)^(3/2) * dq
dFy = kQ^2/a*y/(y^2+x^2)^(3/2)*dy
Fy = int(0, a, kQ^2/a*y/(y^2+x^2)^(3/2)*dy)
Fy = kQ^2/a int(0, a, dy y/(y^2+x^2)^(3/2))
Fy = kQ^2/a * (1/x - 1/(a^2+x^2)^(1/2))

Answer 2

The x and y components of the E-field produced by the charge distribution Q at points on the positive x axis is; $E = \frac{\lambda*d\theta}{4\pi \epsilon_{0} r_{0} }(sin\theta i + (cos\theta + 1)j)$

What is the magnitude of magnetic field?

Let the linear charge density = λ

Let the length of the rod is l and the total charge distributed on the rod is Q. Thus;

$dE = \frac{dQ}{4\pi \epsilon_{0} r^{2} }$

Thus;

$dE = \frac{\lambda*da}{4\pi \epsilon_{0} x^{2} }$

cos θ = $\frac{r_{0} }{r}$

r = r₀ * sec θ

$dE = \frac{\lambda*da}{4\pi \epsilon_{0} r_{0} ^{2} sec^{2}\theta }$

taking x component and y component gives;

$dE = \frac{\lambda*d\theta}{4\pi \epsilon_{0} r_{0} }(cos\theta i + sin\thetaj)$

Integrating both sides gives;

$\int\limits^E_0 {E}= \frac{\lambda}{4\pi \epsilon_{0} r_{0} }(\int\limits^\theta _0 { cos\theta d\theta i + \int\limits^\theta _0 sin\theta d\theta j)$

⇒ $E = \frac{\lambda*d\theta}{4\pi \epsilon_{0} r_{0} }(sin\theta i + (cos\theta + 1)j)$

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