Question

Point charge are placed the corners of square of side 'a'. The value and nature of q can be best represented, to make the net force on Q zero as–

• A. $- 2\sqrt{2}$Q
• B. $2\sqrt{2}Q$
• C. $\frac{- Q}{2\sqrt{2}}$
• D. $\frac{Q}{2\sqrt{2}}$

Answer 1

Answer: (C)

$\frac{- Q}{2\sqrt{2}}$

Answer 2

Resultant of F and F = F'' =

2F cos $\frac { 90 } { 2 }$= $\sqrt { 2 }$ $\frac { \mathrm { kQq } } { \mathrm { a } ^ { 2 } }$

To make force of Q zero :

F' + F'' = 0 Ž + $\frac { \sqrt { 2 } \mathrm { kQq } } { \mathrm { a } ^ { 2 } }$ = 0