Question
Four charges equal to $$ - Q$$ are placed at the four corners of a square and a charge $$q$$ is at its centre. If the system is in equilibrium the value of $$q$$ is
A.
$$ - \frac{Q}{2}\left( {1 + 2\sqrt 2 } \right)$$
B.
$$\frac{Q}{4}\left( {1 + 2\sqrt 2 } \right)$$
C.
$$ - \frac{Q}{4}\left( {1 + 2\sqrt 2 } \right)$$
D.
$$\frac{Q}{2}\left( {1 + 2\sqrt 2 } \right)$$
Answer :
$$\frac{Q}{4}\left( {1 + 2\sqrt 2 } \right)$$
Solution :
Net field at $$A$$ should be zero
$$\eqalign{ & \sqrt 2 {E_1} + {E_2} = {E_3} \cr & \therefore \frac{{kQ \times \sqrt 2 }}{{{a^2}}} + \frac{{kQ}}{{{{\left( {\sqrt 2 a} \right)}^2}}} = \frac{{kq}}{{{{\left( {\frac{a}{{\sqrt 2 }}} \right)}^2}}} \cr} $$
$$ \Rightarrow \frac{{Q\sqrt 2 }}{1} + \frac{Q}{2} = 2q \Rightarrow q = \frac{Q}{4}\left( {2\sqrt 2 + 1} \right)$$
Net field at $$A$$ should be zero
$$\eqalign{ & \sqrt 2 {E_1} + {E_2} = {E_3} \cr & \therefore \frac{{kQ \times \sqrt 2 }}{{{a^2}}} + \frac{{kQ}}{{{{\left( {\sqrt 2 a} \right)}^2}}} = \frac{{kq}}{{{{\left( {\frac{a}{{\sqrt 2 }}} \right)}^2}}} \cr} $$
$$ \Rightarrow \frac{{Q\sqrt 2 }}{1} + \frac{Q}{2} = 2q \Rightarrow q = \frac{Q}{4}\left( {2\sqrt 2 + 1} \right)$$
