Question

Three point charges $Q,4Q$ and $16Q$ are placed on a straight line $9cm$ long. Charges are placed in such a way that the system has minimum potential energy. Then

• A. $4Q$ and $16Q$ must be at the ends and $Q$ at a distance of $3cm$ from $16Q$
• B. $4Q$ and $16Q$ must be at the ends and $Q$ at a distance of $6cm$ from $16Q$
• C. Electric field at the position of $Q$ is zero
• D. Electric field at the position of $Q$ is $\frac{Q}{4\pi {\epsilon }_0}$

Answer 1

Answer: (B), (C)

$4Q$ and $16Q$ must be at the ends and $Q$ at a distance of $6cm$ from $16Q$
Electric field at the position of $Q$ is zero

first of all we have to place the two higher charges at the ends so that potential energy between them which contributes maximum in the total potential energy will be minimum.

Now we have to find the position of charge $Q$ such that potential energy of the system will be minimum.

let $Q$ is at a distance $x$ from $16Q$

therefore potential energy of $Q$ will be

$U=k\frac{16Q^2}{x}+k\frac{4Q^2}{d-x}$

$d$ is length of the line.

Now we have to minimize U therefore

$dU/dx=-k\frac{16Q^2}{x^2}+k\frac{4Q^2}{{(d-x)}^2}=0$

$\frac{4}{x^2}=\frac{1}{{(d-x)}^2}$

which gives $x=2d/3$

$x=6cm$

we can also see that for

$x<6$ ; $dU/dx<0$

$x>6$ ; $dU/dx>0$

therefore potential energy at $x=6cm$ is minimum

since at $x=6cm$

$dU/dx=0$ therefore force on charge $Q$ which is given by

$F=-\frac{dU}{dx}=0$ at $x=6cm$

Therefore electric field at $x=6cm$ is

$E=F/Q=0$

therefore option (B) and (C) are correct