Question

Two free charges q and 4q are placed at a distance d apart. A third charge Q is placed between them at a distance x from charges q such that the system is in equilibrium.Then

• A. $Q=\frac{-4q}{9}$
• B. $Q=\frac{7q}{9}$
• C. $Q=-\frac{3q}{9};$
• D. $Q=-\frac{2q}{9}$

Answer 1

The correct option is (A) $Q=\frac{-4q}{9}$

Step 1, Given data

Charges $=q$ and $4q$ distance $=d$

third charge $=Q$

distance from charge q $=x$

Step 2, Finding the value of Q

We can write

Since the charges are in equilibrium

So,

charges $=q$

$\frac{Kq}{x^2}=\frac{K\left(4q\right)}{{(L-x)}^2}$

or, $x=\frac{L}{3}⟶\left(1\right)$

This is the position where the third charge should be placed. Now, we have to determine the nature and magnitude of it.

For this just try to satisfy the condition of equilibrium on any one of the remaining two charges. Let's take $+q$ charges. The force on it due to $+4q$ charge is directed leftwards so the force on it due to the third charge should be directed rightwards. This implies that the force between the $+q$ charge and the third charge should be of attractive nature. Thus the nature of the $3rd$ charge is $(-ve)$.

Taking the third charge to be $-Q$ (say) and then on applying the condition of equilibrium on $+q$ charge

Forces will be equal

So,

$\frac{KQ}{x^2}=\frac{-K\left(4q\right)}{L^2}$

Putting value of x from equation 1

$\frac{KQ}{{\left(L/3\right)}^2}=\frac{-K\left(4q\right)}{L^2}$

Or,

$\Rightarrow \frac{9KQ}{L^2}=\frac{-4Kq}{L^2}$

$\Rightarrow 9Q=-4q$

Or, $Q=\frac{-4q}{9}$

Hence the value of Q is $Q=\frac{-4q}{9}$