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Electrostatics question

2019 · 9 Jan · Shift 1 · Q68

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2019 · 9 Jan · Shift 1 · Q68

Electrostatics question

2019 · 9 Jan · Shift 1 · Q68

JEE MainPhysicsElectrostaticsMCQ+4 / −1

Three charges + Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is :

A
$-{Q \over 4}$
B
+ ${Q \over 2}$
C
+ ${Q \over 4}$
D
$-{Q \over 2}$

View written solutionFree

Correct answer: A

Understand the configuration

Charges are placed on the $x$ -axis at:

$+Q$ at $x=0$
$q$ at $x=\dfrac d2$
$+Q$ at $x=d$

We need the net force on the charge $+Q$ at the origin to be zero.

Force on $+Q$ at $x=0$ due to $+Q$ at $x=d$

Both charges are positive, so the force is repulsive. Thus, the charge at $x=d$ pushes the charge at the origin toward the left.

Magnitude:

F_1 = k\frac{Q\cdot Q}{d^2} = k\frac{Q^2}{d^2}

Force on $+Q$ at $x=0$ due to charge $q$ at $x=\dfrac d2$

Distance between them is $\dfrac d2$ . Magnitude of force:

F_2 = k\frac{Q|q|}{(d/2)^2} = k\frac{Q|q|}{d^2/4} = 4k\frac{Q|q|}{d^2}

For the net force on the origin charge to be zero, this force must act toward the right to balance $F_1$ .

Since the charge at $x=\dfrac d2$ is to the right of the origin, it will pull the positive charge at origin to the right only if $q$ is negative.

So, $q<0$ .

Equate magnitudes for zero net force

F_2 = F_1

4k\frac{Q|q|}{d^2} = k\frac{Q^2}{d^2}

Cancel $k$ , $Q$ , and $d^2$ :

4|q| = Q

|q| = \frac Q4

Since $q$ is negative,

q = -\frac Q4

Check options

A: $-\dfrac Q4$ ✅
B: $+\dfrac Q2$ ❌
C: $+\dfrac Q4$ ❌
D: $-\dfrac Q2$ ❌

Therefore, the correct option is A.

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