Question

A charge of 6.65 mC is placed at each corner of a square 0.500 m on a side.

Answer 1

Answer:

Explanation:

Given that:

length of side , a = 0.5 m

charge , q = 6.65 mC

length of diagonal , d = 0.5 * sqrt(2)

d = 0.707 m

F is the force due to adjacent particle ,

F1 is the force due to diagonal particle

Now , for the net charge on a particle

Fnet = 2 * F * cos(45) + F1

Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2

Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)

Fnet = 3.05 *10^6 N

the magnitude of net force acting on each particle is 3.05 *10^6 N

part B)

for the direction of particle

d) along the line between the charge and the center of the square outward of the center