Question

four point charges Q, q, Q, q are placed at the corners of a square of side 'a'as shown in the figure .Find the resultant electric force on a charge Q at C.​

Answer 1

Given a square ABCD of sides 'a'. Four charges Q, q, Q and q are placed at A, B, C and D as shown in fig.

Also we can see that, F1 and F2 are perpendicular to each other and Fr is the resultant force.

Now,

$\sf{F=\dfrac{1}{4\pi \epsilon_{o} }. \dfrac{q1q2}{ {r}^{2}}}$

q1 = q, q2 = Q and r = a

$\sf{F_1=\dfrac{1}{4\pi \epsilon_{o} }. \dfrac{qQ}{ {a}^{2}}}$..........(1st equation)

Similarly,

$\sf{F_2=\dfrac{1}{4\pi \epsilon_{o} }. \dfrac{qQ}{ {a}^{2}}}$..........(2nd equation)

So, we can say that F1 = F2. Now, as both of them i.e. F1 and F2 are perpendicular to each other. So,

F' i.e. Resultant force is given by

$\sf{F'=\sqrt{F_{1}^{2}+F_{2}^{2}+2\;F_{1}\;F_{2}\;\cos90^{\circ}}}$

Let's take a one more assumption that F1 = F2 = M

$\sf{F'=\sqrt{M{^2}+M{^2}+2M(M)(0)}}$

$\sf{F'=\sqrt{M{^2}+M{^2}}}$

$\sf{F'=\sqrt{2M{^2}}}$

$\sf{F'=M\sqrt{2}}$

OR

$\sf{F'=\sqrt{2}.\dfrac{1}{4\pi \epsilon_{o} }. \dfrac{qQ}{ {a}^{2}}}$

Also, Resultant force of F1 and F2 is Fr.

F" that is net force is equal to sum of Fr and F'.

F" = Fr + F'

$\sf{F_r=\dfrac{1}{4\pi \epsilon_{o} }. \dfrac{QQ}{ {(\sqrt{2a})}^{2}}}$

So,

$\sf{F"=\dfrac{1}{4\pi \epsilon_{o} }. \dfrac{QQ}{ {(\sqrt{2a})}^{2}}+\sqrt{2}.\dfrac{1}{4\pi \epsilon_{o} }. \dfrac{qQ}{ {a}^{2}}}$

$\sf{F"=\dfrac{Q}{4\pi \epsilon_{o}}\bigg(\dfrac{\sqrt{2q}}{a^2}+\dfrac{Q}{2a^2}}\bigg)$

$\sf{F"=\dfrac{Q}{4\pi \epsilon_{o}}\bigg(\dfrac{2\sqrt{2q}+Q}{2a^2}}\bigg)$