Question

four charges are placed at the vertices of a square of side a,calculate force acting on any one of the charges​

Answer 1

Given:

Four charges are placed at the vertices of a square of side a.

To find:

Force on any one charge?

Calculation:

• Let's assume that all charges are equal.

Now, force in q due to charge 1 :

$F_{1} = \dfrac{k {q}^{2} }{ {a}^{2} }$

Similarly force on q due to charge 2:

$F_{2} = \dfrac{k {q}^{2} }{ {(a \sqrt{2} )}^{2} }$

$\implies F_{2} = \dfrac{k {q}^{2}}{2{a}^{2} }$

Now, force on q due to charge 3:

$F_{3} = \dfrac{k {q}^{2} }{ {a}^{2} }$

Now, net force due to charge 1 and 3:

$F_{1,3} = \sqrt{{(\dfrac{k {q}^{2} }{ {a}^{2} } )}^{2} + {(\dfrac{k {q}^{2} }{ {a}^{2} } )}^{2} }$

$F_{1,3} = \dfrac{ \sqrt{2} k {q}^{2} }{ {a}^{2} }$

Now, total force on charge q will be :

$F = F_{2} + F_{1,3}$

$\implies F = \dfrac{k {q}^{2} }{2 {a}^{2} } + \dfrac{ \sqrt{2}k {q}^{2} }{ {a}^{2} }$

$\implies F = \dfrac{k {q}^{2} }{ {a}^{2} } ( \dfrac{1}{2} + \sqrt{2} )$

$\implies F = \dfrac{k {q}^{2} }{ {a}^{2} } ( \dfrac{ 2\sqrt{2} + 1 }{2} )$

So, final answer is:

$\boxed{ \bf F = \dfrac{k {q}^{2} }{ {a}^{2} } ( \dfrac{ 2\sqrt{2} + 1 }{2} )}$