Assertion: Consider two identical charges pla | vedclass

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(A) Let two identical charges $Q$ be placed at $x = -d$ and $x = +d$. The midpoint $O$ is at $x = 0$.If a positive test charge $q$ is placed at $O$,th

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If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Vedclass · Answer

(A) Let two identical charges $Q$ be placed at $x = -d$ and $x = +d$. The midpoint $O$ is at $x = 0$.If a positive test charge $q$ is placed at $O$,the force from the left charge is $F_1 = \frac{kQq}{d^2}$ (towards right) and the force from the right charge is $F_2 = \frac{kQq}{d^2}$ (towards left). The net force is $F_{net} = F_1 - F_2 = 0$. Thus,the charge is in equilibrium.If the test charge $q$ is displaced by a small distance $x$ towards the right,the force from the right charge increases and the force from the left charge decreases. The net force will act towards the left (opposite to the displacement),tending to restore the charge to its original position. This confirms stable equilibrium for displacements along the $x-$axis.Since the Assertion is correct and the Reason correctly states that the net force is zero (which is a condition for equilibrium),the Reason is the correct explanation for the Assertion.

Assertion: Consider two identical charges placed at a distance $2d$ apart along the $x-$axis. The equilibrium of a positive test charge placed at the point $O$ midway between them is stable for displacements along the $x-$axis.
Reason: The net force on the test charge at point $O$ is zero.

Similar Questions

Two identical charges $Q$ are separated by a distance $r$. $A$ third charge $q$ is placed at the midpoint of the line joining the two charges such that all three charges are in equilibrium. In this state,$q = $ . . . . . . .

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