Question

Two point charges of charge value Q and q are placed at a distance of x and x/2 respectively from a third charge of charge value 4q, all charges being in the same straight line. Calculate the magnitude and nature of charge Q, such that net force experienced by the charge q is zero. (Assuming q is located between charges 4q and Q).

Answer 1

Hint: Force experienced by a test charge q1 due to the other charge q2 present at a distance r from it is given by:
 $\Rightarrow F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} $
Where k is constant.
In this question we calculate all the forces acting on charge q due to Q and 4q and then taking their scalar sum and equating it to zero as net force on q should be zero.

Complete step by step solution

Distance between Q and q = x – x/2 = x/2
Let the force experienced by charge q due to Q be FQ and that due to 4q be F4q.
 $ \begin{align}
 &\Rightarrow {{F}_{Q}}=k\dfrac{qQ}{{{\left( {}^{x}/{}_{2} \right)}^{2}}} \\
 &\Rightarrow {{F}_{Q}}=k\dfrac{4qQ}{{{x}^{2}}} \\
\end{align} $
Also
 $ \begin{align}
 &\Rightarrow {{F}_{4q}}=k\dfrac{q(4q)}{{{\left( {}^{x}/{}_{2} \right)}^{2}}} \\
 &\Rightarrow {{F}_{4q}}=k\dfrac{16{{q}^{2}}}{{{x}^{2}}} \\
\end{align} $
 $ \begin{align}
 &\Rightarrow {{F}_{net}}={{F}_{Q}}+{{F}_{4q}} \\
 &\Rightarrow 0=k\dfrac{4qQ}{{{x}^{2}}}+k\dfrac{16{{q}^{2}}}{{{x}^{2}}} \\
 &\Rightarrow k\dfrac{4qQ}{{{x}^{2}}}=-k\dfrac{16{{q}^{2}}}{{{x}^{2}}} \\
 &\Rightarrow 4Q=-16q \\
 &\Rightarrow Q=-4q \\
\end{align} $
Therefore, the magnitude of charge Q is 4q which is negative in nature.

Note
We can avoid this whole calculation for the value of Q by noticing that charge q is placed at the center of 4q and Q where net force experienced is to be zero which means that it should be surrounded by equal and opposite charges to make the net force experienced as zero. Thus if charge on one of its sides is 4q then the other charge should be -4q which gives us the value of charge Q without any calculation.