Part A: Calculate the magnitude of the electric field at one corner of a square
ID: 1410518 • Letter: P
Question
Part A: Calculate the magnitude of the electric field at one corner of a square 1.52 m on a side if the other three corners are occupied by 3.75×106 C charges.
Part B:What is the direction of the electric field at the corner?
Answer choices:
(A)along the line between the corner and the center of the square outward of the center
(B)along the side of the square between the corner and one of the charges toward the charge
(C)along the side of the square between the corner and one of the charges outward of the charge
(D)along the line between the corner and the center of the square toward the center
Explanation / Answer
Electric field is defined as the force experienced by a unit positive charge. Suppose a square ABCD is given where the charges are at B, C and D and electric field is to be worked out at A.
Electric field due to the charge B and D would be equal in magnitude
= kq/r^2 = (9x10^9) * (3.75 x 10^-6) / (1.52)^2 N/C = 1.69 x 10^4
The resultant of two such charges will be along CA
= 2cos45° * [1.69 x 10^4] N/C = 23900 N/C
Electric field due to charge at C will be along CA
= kq/(2r)^2 = (9x10^9) * (3.75 x 10^-6) / (1.522)^2 = 8450 N
Total field = 23900 + 8450 = 32350 N/C in the direction CA.