Zero potential spots Charges of -q and +2q are fixed in place, with a distance d
ID: 2164974 • Letter: Z
Question
Zero potential spotsCharges of -q and +2q are fixed in place, with a distance d between them (see the drawing). A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is at least one spot where the total potential is zero.
(a) At this spot, is the magnitude of the potential from the positive charge greater than, less than, or equal to the magnitude of the potential from the negative charge?
greater than
less than
equal to
Incorrect: Your answer is incorrect.
Incorrect. The potential from the positive charge is positive, while the potential from the negative charge is negative. At the spot in question, think about how the magnitudes of these potentials must be related so that the total potential is zero.
(b) Is the distance from the positive charge to the zero-potential spot greater than, less than, or equal to L?
The distance is less than L.
The distance is greater than L.
The distance is equal to L.
Correct: Your answer is correct.
Correct. At the zero-potential spot, the magnitude of the potential produced by -q must be equal to that produced by +2q. The potential produced by a point charge is directly proportional to the charge and inversely proportional to the distance (see Equation 19.6 in ). Thus, the distance between the +2q charge and the zero-potential spot must be greater than L, which is the distance between the -q charge and the zero-potential spot.
(c) How many spots on the dashed line are there where the total potential is zero?
There is only one spot.
There are two spots.
There are three spots.
Correct: Your answer is correct.
Correct. There are two spots on the dashed line where the total potential is zero. One spot is at a distance L above the negative charge. By symmetry, the other spot is at a distance L below the negative charge.
Charges of -q and +2q are fixed in place, with a distance d = 2.45 m between them. A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is a spot where the total potential is zero.
(d) What is the algebraic expression for the total potential VTotal at a given spot (not necessarily the spot where the potential is zero) that is produced by two point charges q1 and q2? Express your answer in terms of q2 and q2, the distance r1 from the charge q1 to the spot, the distance r2 from the charge q2 to the spot, and the constant k. (Answer using q_1 for q1, q_2 for q2, r_1 for r1, r_2 for r2, and k.)
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(e) Find the distance L.
Number Unit
L = Incorrect: Your answer is incorrect.
Your response differs from the correct answer by more than 10%. Double check your calculations. Correct: Your answer is correct.
part e is the only thing i need, just thought it would be best to show the whole problem
Explanation / Answer
Let the charges lie on X axis, with qA on the left. At the required point (or points) P, the electric field due to qA is EA and that due to qB is EB. EA and EB must have opposite directions and same magnitude. The point(s) must also on the X axis. There are 3 regions on the axis: 1) left of qA 2) between qA and qB 3) right of qB In region 1, no such P because ¦qA¦>qB and the distance PA EB. In region 2, no such P because EA and EB have same direction. In region 3, there's such P. Suppose the distance PB = r, then PA = r + d ¦EA¦ = K¦qA¦/(r+d)^2 EB = KqB/r^2 ¦EA¦= EB, ¦qA¦/(r+d)^2 = qB/r^2 (r+d)^2/r^2 = ¦qA¦/qB (r+d)/r = sqrt(¦qA¦/qB) d/r = sqrt(¦qA¦/qB) - 1 so r = d/[sqrt(¦qA¦/qB) - 1] = 1.72d Therefore the required point P is located on the line AB, with the distance 1.72d from qB and 2.72d from qA.Related Questions
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