Consider a square which is 1.0 m on a side. Charges are placed at the corners (m
ID: 2279498 • Letter: C
Question
Consider a square which is 1.0 m on a side. Charges are placed at the corners (marked A,B,C D) of the square as shown in the figure.(Figure 1) Q = 5 ?C and q = 2 ?C The distance of each corner to the center of the square is 0.707 m.
I have no idea really how to go about this, so if anyone could explain how that would be great!
Explanation / Answer
a)
At the center, electric field due to charges at A & C will cancel out each other due to having equal and opposite field direction.
So, resultant E will only be due to charges at B & D, i.e.
E = kq / r^2 + k |-q| / r^2
= k x 2q / r^2
= 8.98 x 10^9 x 2 x 2 x 10^-6 / (0.71^2)
= 71.26 x 10^3 N/C.
b)
Electric field at the center will be in the direction along the line joining B & D towards D.
Angle = 90 + 45 = 135 degree (Since, it is a square)
c)
Electric potential, V = k x (2Q + q - q) / r
= 8.98 x 10^9 x (2 x 5 + 2 - 2) x 10^-6 / 0.71
= 12.64 x 10^4 J/C.
d)
Force acting on the charge will be E x (q) in the direction along the line joining B & D towards B as the charge on which the force is acting is negative.
So, F = 71.26 x 10^3 x 1 x 10^-6
= 71.26 x 10^-3 N
Fx = F x Cos 45
= 50.38 x 10^-3 N. in positive x axis.
e)
Fy = F Sin 45
= 50.38 x 10^-3 N. in negative y axis.
f)
Required Work to be done = change in electrostatic potential energy
= PEi - PEf
= V x q - 0 (PEf = 0 because at infinity, the energy on the charge will be zero.)
= 12.64 x 10^4 x 1 x 10^-6
= 0.1264 J.