Question

A charge of 1.20 mC is placed at each corner of a square 0.100 m on a side.

Part A

Determine the magnitude of the force on each charge.

Part B

Determine the direction of the force on each charge.

Answer 1

A)

Refer the above figure,

Here d= 0.1m and Q=1.2mC

F41= kq1q4/d^2 = kq^2/d^2   

F41x= F41cos0= F41= kq^2/d^2   

F41y= F41sin0=0

F43 = kq3q4/d^2 = kq^2/d^2

F43x= F41cos90= 0

F43y= F41sin90= F41= kq^2/d^2   

F42 = kq2q4/d^2 = kq^2/(sqr2d)^2 = kq^2/(2d^2)

F42x= F42cos45 = kq^2/(2d^2)*1/sqrt2 = kq^2/(2sqrt2d^2)

F42y= F42cos45 = kq^2/(2d^2)*1/sqrt2 = kq^2/(2sqrt2d^2)

F4x= F41x+F42x+F43x = kq^2/d^2   + kq^2/(2sqrt2d^2) + 0 = kq^2/d^2 *(1+1/2sqr2)

F4y= F41y+F42y+F43y = 0+ kq^2/(2sqrt2d^2) + kq^2/d^2 = kq^2/d^2 *(1+1/2sqr2)

F4= sqrt(F4x^2+F4y^2) = kq^2/d^2*(sqrt2+1/2)=

Plugging values,

F4= (9*10^9*(6*10^-3)^2/(0.1^2))*(sqr2+1/2) = 2.5*10^6 N

B) ? = tan^-1(F4y/F4x) = tan^-1(1) = 45 deg

For each charge, the net force will be the magnitude determined above, and will lie along the line from the center of the square out towards the charge.