Question

Open in App

Verified by Toppr

Correct option is D)

** **

$Hint:$ **Let distance
between $A$ and $B$ be $d$.**

$Step 1:$ F**ind potential difference of both**

Let distance between $A$ and $B$ be $d$

Then potential at any distance $x$ from point $A$ is given by

$V=xkq +(d−x)kq $

$V=kq(x1 +(d−x)1 )$

NoW

$Step 2:$** Differentiating
between both potential energy**

$dV/dx=(x(d−x))_{2}kqd(2x−d) $

therefore for $x<d/2$

$dV/dx<0$ therefore potential decreases and for $x>d/2$

**$dV/dx>0$ therefore potential increases Hence first decreases then
increases option (D)**

Solve any question of Electrostatic Potential and Capacitance with:-

0

0