Two positive charges Q and 4Q are placed at p | vedclass

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Question

(A) The total electric potential $V$ at a point $P$ between the charges is given by $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{r} + \frac{4Q}{d-

Vedclass · Accepted Answer

$\frac{d}{3}$ units to the right of $A$

Vedclass · Answer

(A) The total electric potential $V$ at a point $P$ between the charges is given by $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{r} + \frac{4Q}{d-r} \right)$, where $r$ is the distance of $P$ from $A$.For the potential to be minimum, we set $\frac{dV}{dr} = 0$.$\frac{dV}{dr} = \frac{1}{4\pi\varepsilon_0} \left( -\frac{Q}{r^2} + \frac{4Q}{(d-r)^2} \right) = 0$.This implies $\frac{Q}{r^2} = \frac{4Q}{(d-r)^2}$, which is equivalent to the condition where the net electric field $E = 0$.Taking the square root on both sides: $\frac{1}{r} = \frac{2}{d-r}$.Solving for $r$: $d - r = 2r \Rightarrow 3r = d \Rightarrow r = \frac{d}{3}$.Thus, the potential is minimum at a distance of $\frac{d}{3}$ units to the right of $A$.

Two positive charges $Q$ and $4Q$ are placed at points $A$ and $B$ respectively, where $B$ is at a distance $d$ units to the right of $A$. The total electric potential due to these charges is minimum at point $P$ on the line segment joining $A$ and $B$. What is the distance of $P$ from $A$?

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