Problem 19.52 The electric field at the point x =5.00cm and y =0 points in the p
ID: 581577 • Letter: P
Question
Problem 19.52
The electric field at the point x=5.00cm and y=0 points in the positive x direction with a magnitude of 9.00 N/C . At the point x=10.0cm and y=0 the electric field points in the positive x direction with a magnitude of 19.0 N/C . Assume this electric field is produced by a single point charge.
Part A
Find the charge's location.
SubmitMy AnswersGive Up
Part B
Find the sign of the charge.
SubmitMy AnswersGive Up
Part C
Find the magnitude of the charge.
SubmitMy AnswersGive Up
Problem 19.52
The electric field at the point x=5.00cm and y=0 points in the positive x direction with a magnitude of 9.00 N/C . At the point x=10.0cm and y=0 the electric field points in the positive x direction with a magnitude of 19.0 N/C . Assume this electric field is produced by a single point charge.
Part A
Find the charge's location.
x = cmSubmitMy AnswersGive Up
Part B
Find the sign of the charge.
Find the sign of the charge. positive negativeSubmitMy AnswersGive Up
Part C
Find the magnitude of the charge.
q = pCSubmitMy AnswersGive Up
Explanation / Answer
Note that the electric field E at A(5.00, 0) and B(10.0, 0) are both in the +X direction.
If the charge q is positive, it must be at the left of A and the distance Aq < Bq. But it would produce EA > EB. This contradicts the condition in the question.
If the charge q is negative, it must be at the right of B, the distance Aq > Bq and gives EA < EB. This is the case.
So the charge is negative. Let's use q to denote the magnitude of the charge and its position (x, 0).
EA = Kq/(x-5)^2 = 9 (1)
EB = Kq/(x-10)^2 = 19 (2)
(1)/(2): (x-10)^2/(x-5)^2 = 9/19
(x-10)/(x-5) = sqrt(9/19)
1 - (x-10)/(x-5) = 1 - sqrt(9/19)
5/(x-5) = 1 - sqrt(9/19)
x = 5 + 5/[1 - sqrt(9/19)] = 20.6 cm = 0.206 m
From (1), q = 9(x-5)^2/K = 8*0.232^2/(9*10^9) = 9.47 * 10^9 C
Remember this is a positive charge.