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

Question

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

Part A

Determine the magnitude of the force on each charge.

Part B

Determine the direction of the force on a charge.

A) along the side of the square outward of the other charge that lies on the side B) along the side of the square toward the other charge that lies on the side C) along the line between the charge and the center of the square toward the center d) along the line between the charge and the center of the square outward of the center

Explanation / Answer

Here,

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

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