Question

two fixed point charges +4e and +e unit are separated by a distance a.Where the third point charge should be placed from +4e charge for it to be in equilibrium

Answer 1

as in case of equilibrium,
F1= force due to+ e charge
F2= force due to+4 e charge
F1-F2 =0
so f1= f2 further answer in photo

Answer 2

Third charge should be $\bold{\frac{2}{3 a}}$ from the +4e charge.

Given data suggest that there are two fixed point charges as +4e and +e which are at a distance of ‘a’ from each other. So, q1 = +4e and q2 = +e and also distance = a. Let us assume that the third charge be q3 which is at a distance = x from q1.  

So, the distance from q2 will be = a – x.  

$\Rightarrow$ Force on q3 exerted by q1,

$F=\frac{k q 1 q 3}{x^{2}} \Rightarrow \text{and force on q3 exerted by q2}, F^{\prime}=\frac{k q 2 q 3}{(a-x)^{2}}$

Now the given situation is that the three are in equilibrium so the force exerted on q3 must be balanced.  

$\begin{array}{l}{\Rightarrow \mathrm{F}=\mathrm{F}^{\prime}} \\ \\ {\Rightarrow \frac{k q 1 q 3}{x^{2}}=\frac{k q 2 q 3}{(a-x)^{2}}}\end{array}$

$\begin{array}{l}{\Rightarrow \frac{4 e}{x^{2}}=\frac{e}{(a-x)^{2}}} \\ \\ {\Rightarrow \frac{4}{x^{2}}=\frac{1}{(a-x)^{2}}} \\ \\ {\Rightarrow \frac{2}{x}=\frac{1}{a-x}}\end{array}$

$\begin{array}{l}{\Rightarrow 2(\mathrm{a}-\mathrm{x})=\mathrm{x}} \\ {\Rightarrow 2 \mathrm{a}-2 \mathrm{x}=\mathrm{x}} \\ {\Rightarrow 2 \mathrm{a}=3 \mathrm{x}} \\ {\Rightarrow \mathrm{x}=\frac{2 a}{3}}\end{array}$

So, third charge should be $\frac{2}{3 a}$ from the +4e charge.