Question

Four point charges are fixed in position at the corners of a square, as shown in above figure.

Find out the direction of the total electrostatic force on the charge in the lower right-hand corner of the square?

• A. $\nwarrow$
• B. $\leftarrow$
• C. $\searrow$
• D. $\to$
• E. The electric force on this charge is $0$

Answer 1

Answer: (A)

$\nwarrow$

there are three forces acting on charge in the lower right-hand corner

(i)The direction of attraction force$F=K{Q}^2/r^2$

on charge in lower right hand corner by charge in upper right-hand corner will be $\uparrow$

(towards upper right -hand corner)

(ii)The direction of attraction force$F^{\prime }=K{Q}^2/r^2$

on charge in lower right hand corner by charge in lower left-hand corner will be $\leftarrow$

( towards lower left -hand corner)

(iii)The direction of repulsion force$F^{\prime \prime }=K{Q}^2/{2r}^2$

on charge in lower right hand corner by charge in upper left-hand corner will be $\searrow$

( towards lower right -hand corner)

the first two forces have equal magnitude because charges are at equal distances from lower right hand corner charge

since F and F' are equal ,their resultant will be equally inclined to them i.e. it will act along $\nwarrow$

their resultant is F'''=$\sqrt{F{}^2+{F^{\prime }}^2}=\sqrt{2}F=\sqrt{2}K{Q}^2/r^2$ along $\nwarrow$ ( they are equal in magnitude)

now we have two opposite forces F'' and F''' as the magnitude of F'' is greater than F'' therefore resultant direction will be towards the upper left-hand corner $\nwarrow$.