Question 14 Marks

Draw a graph showing the change in potential corresponding to the change in distance from a point charge. For the charged system shown in the figure, prove that the potential difference between points $A$ and $B$ is

Answer

In the figure, the potential at point A due to $+q$ and $-q$ charges respectively is

$V _1=\frac{ l }{4 \pi \in_0} \frac{q}{a}$
And $\quad V _2=\frac{1}{4 \pi \in_0} \frac{(-q)}{(a+d)}$
Therefore the net potential at point A
$V_A=V_1+V_2$
$=\frac{1}{4 \pi \in_0}\left(\frac{q}{a}\right)+\frac{1}{4 \pi \in_0}\left(\frac{-q}{a+d}\right)$
$V _{ A }=\frac{1}{4 \pi \in_0}\left(\frac{1}{a}-\frac{1}{a+d}\right)$
$V _{ A }=\frac{q}{4 \pi \in_0}\left(\frac{a+d-a}{a(a+d)}\right)=\frac{1}{4 \pi \in_0} \frac{q d}{a(a+d)}$
Similarly net potential at point B
$V _{ B }=\frac{q}{4 \pi \in_0}\left(\frac{1}{a+d}-\frac{1}{a}\right)$
$=\frac{-1}{4 \pi \in_0} \frac{q d}{a(a+d)}$
Therefore,
$V_A-V_B$$\frac{1}{4 \pi \in_0} \frac{q d}{a(a+d)}-\left(\frac{-1}{4 \pi \in_0} \frac{q d}{a(a+d)}\right)$
$\begin{aligned} V _{ A }- V _{ B } & =\frac{1}{4 \pi \in_0} \frac{q d}{a(a+d)}[1+1] \\ & =\frac{2 q d}{4 \pi \in_0 a(a+d)}\end{aligned}$
$V _{ A }- V _{ B }=\frac{q}{2 \pi \in_0 a} \cdot \frac{d}{a+d}$

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