Question

Positive charge Q is distributed uniformly along the x-axis from x = 0 to x = a. A positive point charge q is located on the positive x-axis at x = a + r, where r is the distance to the right end of Q.(a) Calculate the x-component of the electric field produced by the charge distribution Q at points on the positive x-axis where x > a.(b) Calculate the y-component of the electric field produced by the charge distribution Q at points on the positive x-axis where x > a.(c) Calculate the magnitude of the force that the charge distribution Q exerts on q.(d) Calculate the direction of the force that the charge distribution Q exerts on q.

Answer 1

Answer:

(a): $k\dfrac{Q}{r(a+r)}.$.

(b): 0

(c):$k\dfrac{qQ}{r(a+r)}.$.

(d): Along positive x axis.

Explanation:

Given that the charge is distributed uniformly on the rod, extended from $x=0$ to $x=a$ and having total charge $Q$.

Let $\lambda$ be the linear charge density of the rod, such that,$\lambda = \dfrac Qa.$

The electric field due to a charge $q$ at a point $r$ distance away is given by

$E=\dfrac{kq}{r^2}$

where $k$ is the Coulomb's constant whose value is $9\times 10^9] Nm^2/C^2$.

Now consider a small line element of the rod of length $dx$ at distance $r$ from $x =a+r$.

The electric field at point $x = a + r$ due to this element is given by

$dE = \dfrac{k\ dq}{x^2}$

$dq$ is the charge on the line charge, then,

$\lambda = \dfrac{dq}{dx}\\\Rightarrow dq=\lambda dx$

Using this value,

$dE = \dfrac{k\lambda\ dx}{x^2}.$

The electric field due to the whole rod is given by

$E=\int dE\\=\int\limits_{r}^{a+r} \dfrac{k\lambda\ dx}{x^2}\\=k\lambda\int\limits^{a+r}_{r} \dfrac{ dx}{x^2}\\=k\lambda\int\limits^{a+r}_{r} x^{-2}\ dx\\=k\lambda \left (-x^{-1} \right )\limits^r_{a+r}\ dt\\=-k\lambda \left ( \dfrac1{a+r}-\dfrac 1r\right ) \\=-k\lambda \left ( \dfrac{r-(a+r)}{r(a+r)}\right ) \\=k\lambda \left ( \dfrac{a}{r(a+r)}\right )\\=k\dfrac{Q}{a} \left ( \dfrac{a}{r(a+r)}\right )\\=k\dfrac{Q}{r(a+r)}.$

This electric is along the x axis only.

(a):

The x component of the electric field at this point = $k\dfrac{Q}{r(a+r)}.$.

(b):

The y component of the electric field at this point = 0.

(c):

The electric field and the electric force are related as

$F=qE.$

The magnitude of the force that this rod exerts on the charge q = $k\dfrac{qQ}{r(a+r)}.$.

(d):

The direction of this force is along the positive x axis.