Question

Four charged particles are held fixed at the corners of a square of side s. All the charges have the same magnitude Q, but two are positive and two are negative. In Arrangement 1, shown above, charges of the same sign are at opposite corners. Express your answers to parts a. and b. in terms of the given quantities and fundamental constants.

a. For Arrangement 1, determine the following.

i. The electrostatic potential at the center of the square

ii. The magnitude of the electric field at the center of the square

The bottom two charged particles are now switched to form Arrangement 2, shown above, in which the positively charged particles are on the left and the negatively charged particles are on the right.

b. For Arrangement 2, determine the following.

i. The electrostatic potential at the center of the square

ii. The magnitude of the electric field at the center of the square

c. In which of the two arrangements would more work be required to remove the particle at the upper right corner from its present position to a distance a long way away from the arrangement?

_________ Arrangement 1

 ___________ Arrangement 2

Justify your answer.

Answer 1

a. 

i. V = ΣkQ/r = k(–Q/r + –Q/r + Q/r + Q/r) = 0

ii. The fields from the charges on opposing corners cancels which gives E = 0

b. 

i. V = ΣkQ/r = k(–Q/r + –Q/r + Q/r + Q/r) = 0

ii. The field from each individual charge points along a diagonal, with an x-component to the right. The vertical components cancel in pairs, and the x-components are equal in magnitude. Each x component being E = kQ/r² cos 45º and the distance from a corner to the center of r² = s²/2 gives

c. Arrangement 1. The force of attraction on the upper right charge is greater in arrangement 1 because the two closest charges are both positive, whereas in arrangement 2 one is positive and one is negative.