Two point charges Q each are placed at a dist | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the two charges $Q$ be placed at points $A$ and $B$ separated by distance $d$. The mid-point is $O$. The charge $q$ is at point $P$ on the per

Vedclass · Accepted Answer

$x=\frac{d}{2\sqrt{2}}$

Vedclass · Answer

(D) Let the two charges $Q$ be placed at points $A$ and $B$ separated by distance $d$. The mid-point is $O$. The charge $q$ is at point $P$ on the perpendicular bisector at distance $x$ from $O$.The distance between each charge $Q$ and charge $q$ is $r = \sqrt{x^2 + (d/2)^2}$.The magnitude of the Coulomb force exerted by each charge $Q$ on $q$ is $F = \frac{kQq}{x^2 + d^2/4}$.The horizontal components of the forces cancel out,and the net force $F_{	ext{net}}$ is directed along the perpendicular bisector:$F_{	ext{net}} = 2F \cos 	heta = 2 \left( \frac{kQq}{x^2 + d^2/4} ight) \left( \frac{x}{\sqrt{x^2 + d^2/4}} ight) = \frac{2kQqx}{(x^2 + d^2/4)^{3/2}}$.To find the maximum force,we set $\frac{dF_{	ext{net}}}{dx} = 0$:$\frac{d}{dx} \left[ 2kQqx (x^2 + d^2/4)^{-3/2} ight] = 0$.Using the product rule: $2kQq \left[ (x^2 + d^2/4)^{-3/2} + x(-3/2)(x^2 + d^2/4)^{-5/2}(2x) ight] = 0$.$(x^2 + d^2/4)^{-3/2} - 3x^2(x^2 + d^2/4)^{-5/2} = 0$.$(x^2 + d^2/4) - 3x^2 = 0 \implies d^2/4 = 2x^2 \implies x^2 = d^2/8$.Therefore,$x = \frac{d}{2\sqrt{2}}$.

Two point charges $Q$ each are placed at a distance $d$ apart. $A$ third point charge $q$ is placed at a distance $x$ from the mid-point on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb force is ...............

Similar Questions

Two identical conducting spheres having unequal positive charges $q_1$ and $q_2$ are separated by a distance $r$. If they are made to touch each other and then separated again to the same distance $r$,the electrostatic force between the spheres in this case will be (neglect induction of charges):

Four charges equal to $-Q$ are placed at the four corners of a square of side length $a$,and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is:

Two point charges $+9e$ and $+e$ are at $16\, cm$ away from each other. Where should another charge $q$ be placed between them so that the system remains in equilibrium?

Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and the distance between them is doubled,then the new force acting on each charge is

Two charges $q$ and $-3q$ are placed fixed on the $x$-axis separated by a distance $d$. Where should a third charge $2q$ be placed such that it will not experience any force?

Vedclass Test Series

Exam Paper Generator

Online Exam Module