Respuesta :
Answer:
Explanation:
Given
two point charges q and -q are placed at corner of square
suppose a is the side length of square
If a charge [tex]q_2[/tex] is placed at a one of the corner then Force on it due to +q
[tex]F_{+q}=\frac{kqq_2}{a^2}[/tex]
Force on [tex]q_2[/tex] due to -q is given by
[tex]F_{-q}=\frac{-kqq_2}{a^2}[/tex]
Net force [tex]F_{net}=\sqrt{(F_{+q})^2+(F_{-q})^2}[/tex]
[tex]F_{net}=\sqrt{2}\frac{kqq_2}{a^2}[/tex]
When charge is placed at center then distance between charges becomes [tex]\frac{\sqrt{2}}{2}=\frac{a}{\sqrt{2}}[/tex]
One charge will cause [tex]q_2[/tex] to move away from it while other attracts it
[tex]F'_{net}=2\times \frac{kqq_2}{(\frac{a}{\sqrt{2}})^2}[/tex]
[tex]F'_{net}=2\times 2\times \frac{kqq_2}{a^2}[/tex]
[tex]F'_{net}=4\frac{kqq_2}{a^2}[/tex]
Therefore When charge is Placed at center then it experience more force
