Question

A line charge of length $l$ and charge $Q$ uniformly distributed over the length is placed at a distance $r$ from one edge from a point charge $q$ as shown. Find the force on the point charge:

A) $\dfrac{{qQ}}{{4\pi {\varepsilon _0}r\left( {r - l} \right)}}$
B) $\dfrac{{qQ}}{{4\pi {\varepsilon _0}\left( {r - \dfrac{1}{l}} \right)}}$
C) $\dfrac{{qQ}}{{4\pi {\varepsilon _0}\left( {\dfrac{1}{{{r^2}}} - \dfrac{1}{{{{\left( {r + l} \right)}^2}}}} \right)}}$
D) None of these.

Answer 1

hint: To solve this type question first we take a small segment of line charge. And assume it as a point charge due to this small segment (small charge) find the force on charge q. In the same manner we calculate force due to every small segment and add them or we can use integration methods to find force due to the whole line charge.We know force between two point charges is given by $F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

Complete step by step solution:
Step 1 : First of all we assume a small segment of length $dx$ on line charge having $x$ distance from point charge $q$
As given length $l$ having charge $Q$ so the linear charge density given as $\lambda = \dfrac{Q}{l}$ means charge per unit length
As we can see in diagram

Segment $dx$ having a small charge can be given by $dQ = \lambda dx$
First we calculate the small force $dF$ on $q$ due to this segment which having charge $dQ$ and distance from $q$ charge is $x$
Apply formula for force between two point charges $F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here ${q_1} = dQ$ and ${q_2} = q$ $r = x$
$\Rightarrow dF = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qdQ}}{{{x^2}}}$
Segment $dx$ having a small charge can be given by $dQ = \lambda dx$
$\Rightarrow dF = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{q\lambda dx}}{{{x^2}}}$
Rearranging it
$\Rightarrow dF = \dfrac{{q\lambda }}{{4\pi {\varepsilon _0}}}\dfrac{{dx}}{{{x^2}}}$
This is the small force on point charge $q$ due to small segment of $dx$

Step 2: If we want to calculate total force by $l$ length charge we have to integrate it
Total force due to $l$ length $Q$ charge $F$ can given as
$\Rightarrow \int {dF} = \dfrac{{q\lambda }}{{4\pi {\varepsilon _0}}}\int\limits_r^{l + r} {\dfrac{{dx}}{{{x^2}}}}$
Net force
$\Rightarrow F = \dfrac{{q\lambda }}{{4\pi {\varepsilon _0}}}\int\limits_r^{l + r} {{x^{ - 2}}dx}$
On integration
$\Rightarrow F = \dfrac{{q\lambda }}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{{x^{ - 1}}}}{{ - 1}}} \right]_r^{l + r}$
$\Rightarrow F = \dfrac{{q\lambda }}{{4\pi {\varepsilon _0}}}\left[ { - \dfrac{1}{x}} \right]_r^{l + r}$
Applying limits
$\Rightarrow F = \dfrac{{q\lambda }}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{1}{r} - \dfrac{1}{{\left( {l + r} \right)}}} \right]$
$\Rightarrow F = \dfrac{{q\lambda }}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{l + r - r}}{{r\left( {l + r} \right)}}} \right]$
$\Rightarrow F = \dfrac{{q\lambda }}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{l}{{r\left( {l + r} \right)}}} \right]$
Now put value of linear charge density given by$\lambda = \dfrac{Q}{l}$
$\Rightarrow F = \dfrac{{qQ}}{{4\pi {\varepsilon _0}l}}\left( {\dfrac{l}{{r\left( {l + r} \right)}}} \right)$
Solving again
$\Rightarrow F = \dfrac{{qQ}}{{4\pi {\varepsilon _0}r\left( {l + r} \right)}}$
So the total force on the point charge $q$ is $\Rightarrow F = \dfrac{{qQ}}{{4\pi {\varepsilon _0}r\left( {l + r} \right)}}$

So option D is correct.

Note: Sometimes students get confused why we use here the limit of integration is $r$ to $\left( {r + l} \right)$. See in diagram the distance between charge and segment of charge vary between $r$ to $\left( {r + l} \right)$ means the closest end of line charge is at distance $r$ and other end of this line charge having $\left( {r + l} \right)$ distance from point charge $q$ .