A particle with a charge of 5.50 nC is placed at the origin of an x y -coordinat

Question

A particle with a charge of 5.50  nC is placed at the origin of an xy-coordinate system, and a charge -2.00  nC is placed on the positive x-axis at 5.00  cm . A third charge 9.00  nC is now placed at the point x=4.00  cm ,y=0.

Part A

Calculate the potential at the point x=4.00  cm , y=0, due to the first two charges. Let the potential be zero far from the charges.

Part B

Calculate the potential at the point x=4.00  cm , y=6.00  cm , due to the first two charges. Let the potential be zero far from the charges.

Part C

If the third charge somehow moves from the point x=4.00  cm , y=0, to the point x=4.00  cm , y=6.00  cm , calculate the work done on it by the field of the first two charges.

Explanation / Answer

let

q1 = 5.5 nc

r1 = distance of q1 from q3 = 0.04 m

q2 = -2 nc

r2 = distance of q2 from q3 = 0.01 m

q3 = 9 nc

part(A)

at x = 4 cm = 0.04 m

potential due to q1 harge = v1 = k*q1/r1 = (9*10^9*5.5*10^-
9)/0.04 = 1237.5 V

potential due to q2 = V2 = k*q2/r2 = -(9*10^9*2*10^-9)/0.01 = -1800 V

total potential V = v1 + v2 = 1237.5-1800 = -562.5 V

---------

part(B)

(x1 , y1) = (0 , 0 )

(x3 , y3 ) = (4 , 6 )

r1 = distance of point (x3 , y3) from q1 = sqrt(4^2+6^2) = 7.21 cm = 0.0721 m

q2 is located at (x2 , y2 ) = 5 , 0

r2 = distance of q2 from (x3 y3 ) = sqrt(x3-x2)^2 + (y3 - y2)^2

r2 = sqrt(1^2+6^2) = 6.08 cm = 0.0608 m

potential due to q1 = k*q1/r1 = (9*10^9*5.5*10^-9)/0.0721 = 686.55 v

potential due to q2 = k*q2/r2 = -(9*10^9*2*10^-9)/0.0608 = -296.06 v

total potential due to 2 charges = V' = v1 + v2 = 390.49 v

----------------

part(C)

work done = dV*q3 = (v-v')*q3

W = -1003.08*8.5*10^-9 = -6.17*10^-6 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.