Question 769: Electrostats & Capacitance

Question

Three charges ' $+3 q$ ', ' $Q$ ' and ' $+q$ ' are placed in a straight line of length ' $l$ ' at points at distances $0, \frac{l}{2}$ and $l$ respectively. The value of Q in order to have the net force on +q to be zero, $\mathrm{Q}=\mathrm{xq}$. The value of $x$ is

Accepted Answer

Three charges on a straight line of length $l$.
At $0$: charge $+3q$.
At $l/2$: charge $Q$.
At $l$: charge $+q$.
Condition: Net force on the charge $+q$ (at $l$) is zero.
We want: $Q = x q$. Find $x$.
Step 1: Forces on the $+q$ (at $x=l$).
Due to $+3q$ at $0$:
Repulsive force, directed to the right.
Magnitude:
$   F_{3q 	o q} = k \cdot \frac{(3q)(q)}{(l-0)^2} = \frac{3k q^2}{l^2}.   $
Due to $Q$ at $l/2$:
Distance: $l - l/2 = l/2$.
Force magnitude:
$   F_{Q 	o q} = k \cdot \frac{|Q| q}{(l/2)^2} = \frac{4 k |Q| q}{l^2}.   $
Direction depends on sign of $Q$:
If $Q$ is positive, then repulsion Ã¢ÂÂ force on $+q$ is to the
right
.
If $Q$ is negative, then attraction Ã¢ÂÂ force on $+q$ is to the
left
.
Step 2: Net force condition.
We want net force on $+q$ to be zero.
Equation in vector form:
$ F_{3q 	o q}\; (	ext{to the right}) \;+\; F_{Q 	o q}\; (	ext{depends on sign}) \;=\; 0. $
So:
$ \frac{3k q^2}{l^2} + \frac{4kQq}{l^2} = 0. $
(Note: The formula automatically accounts for sign of $Q$.)
Step 3: Solve for $Q$.
$ 3q^2 + 4Qq = 0 $
$ 4Qq = -3q^2 $
$ Q = -\frac{3}{4} q. $
Thus,
$ x = -	frac{3}{4}. $
Final Answer:
$ \boxed{	frac{-3}{4}} $
Correct option: B

Answer

$\frac{1}{4}$

Answer

$\frac{-3}{4}$

Answer

-3

Answer

4

Electrostats & Capacitance

Three charges ' +3 q ', ' Q ' and ' +q ' are placed in a straight line of length ' l ' at points at distances 0, \frac{l}{2} and l respectively. The value of Q in order to have the net force on +q to be zero, \mathrm{Q}=\mathrm{xq}. The value of x is

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