Question

ABCD is a square of side "l" .the force at b due to the remaining charges is ..........
i need solution
the diagram is given below
pls answer faaaaaaaaaast

Answer 1

Force acts on charge at B due to charges at A,C and D.

Force between two charges = $\frac{1}{4 \pi e} \frac{q_{1}q_{2} }{r^{2} }$ along the radial direction.

Force on charge at B due to charge at A(Fa) = $\frac{1}{4 \pi e} \frac{q×(-q) }{r^{2} }$
                                                                 = $\frac{1}{4 \pi e} \frac{-q^{2} }{I^{2} }$ (towards A along line AB)

Force on charge at B due to charge at C(Fc) = $\frac{1}{4 \pi e} \frac{q×(-q) }{r^{2} }$
                                                                 = $\frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }$ (towards C along line BC)

Force on charge at D due to charge at A(Fd) = $\frac{1}{4 \pi e } \frac{q×q }{r^{2} }$
                                                                 = $\frac{1}{4 \pi e } \frac{q^{2} }{(√2I)^{2} }$
                                                                 = $\frac{1}{4 \pi e } \frac{q^{2} }{2I^{2} }$ (towards B along the line DB)

Resultant of force acting on B due to A and C (Fac)= √[2×($\frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }$ )²]
                                                                      = √2 × $\frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }$ (towards D along the line DB)

Resultant force (R)= Fac - Fd = √2 × $\frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }$ + $\frac{1}{4 \pi e } \frac{q^{2} }{2I^{2} }$ 

⇒ R = (√2 - 1/2) $\frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }$

⇒R = ($\frac{(2 \sqrt{2} - 1)}{2}$)($\frac{1}{4 \pi e } \frac{-q^{2} }{I^{2} }$)  (towards D along the line BD)

⇒|R| = $\frac{(2 \sqrt{2} - 1) q^{2} }{4 \pi e * 2 I^{2}}$