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 the charge q such that the system is at equilibrium. What is the magnitude of charge and where, that is, at what distance it should be placed?

Answer 1

As a very first step, one could read the question well and hence note down the given important points from it. You could then make a neat diagram for better understanding. Then you could apply the condition for equilibrium in the given system and accordingly find the answer.

Complete step-by-step solution:
In the question, we are given a system where charges q and 4q are placed at d distance apart and another Q is placed at distance x from charge q between them. We are supposed to find this distance x as well as the magnitude of charge Q for the system to be at equilibrium.

For the equilibrium point we have,
$\dfrac{kq}{{{x}^{2}}}=\dfrac{k\left( 4q \right)}{{{\left( d-x \right)}^{2}}}$
$\Rightarrow x=\dfrac{d}{3}$
The charge Q could be kept at this point. Let us assume that the charge Q is negative so as the net force on it could be zero as per the given system. Now, we could apply the condition for equilibrium again to get,
$\dfrac{kQ}{{{x}^{2}}}=\dfrac{k\left( 4q \right)}{{{d}^{2}}}$
$\Rightarrow \dfrac{kQ}{{{\left( \dfrac{d}{3} \right)}^{2}}}=\dfrac{k\left( 4q \right)}{{{d}^{2}}}$
$\Rightarrow 9Q=4q$
$\therefore Q=\dfrac{4q}{9}$
Therefore, we found the magnitude of charge of Q to be $\dfrac{4q}{9}$and also the distance at which it should be kept from charge q to be $\dfrac{d}{3}$.
Note: The constant k that is used in the solution is given by, $k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$. Its value is given by $9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$. The condition for equilibrium is basically derived from the fact that for a charge to be at equilibrium between two charges, the force due each of those charges on the new charge introduced would be 0.