Question

two equal point charges Q is equals to +under root 2 are placed at each of the two opposite corners of a square and eqal point charges q at each of the other two corners . what must be the value of q so that the resultant force in n Q is zero​

Answer 1

Answer:

$-Q/2\sqrt{2}$

Explanation:Each side of square is of length b.

Hence, distance of charge Q from opposite charge Q is PR = $\sqrt{b^{2}+b^{2}  }=\sqrt{2b}$

We can see that if the resultant force on charge Q placed at point R is zero, then charges at Q and S have to be negative.

 

Forces FP, FQ and FS act on the charge Q at point R. Force FP will have vertical and horizontal components.

 Hence, we have

 Fs=Fpcos45

Fq =Fpsin45

Thus, we have

$\frac{1}{4\pi \alpha }*(-q*Q/QR^{2})= 1/4\pi \alpha*(QQ/PR^{2})*COS45 \\\\-q/b^{2} = Q/\sqrt{2b}^{2}*1/\sqrt{2}=Q/2\sqrt{2}b^{2}.\\   q= -Q/2\sqrt{2}$