Two point charges q1=+2.15nC and q2=6.50nC are 0.100 m apart. Point A is midway

Question

Two point charges q1=+2.15nC and q2=6.50nC are 0.100 m apart. Point A is midway between them; point B is 8.00×102 m from q1 and 6.00×102 m from q2. (See the figure below(Figure 1) .) Take the electric potential to be zero at infinity.

Part A

Find the potential at point A.

V = ___V

Part B

Find the potential at point B.

V=____V

Part C

Find the work done by the electric field on a charge of 2.30 nC that travels from point B to point A.

W = _____J

ONLY FIND PART C

The Other answers (A and B ) can be seen here :

http://www.chegg.com/homework-help/questions-and-answers/two-point-charges-q1-215nc-q2-650nc-0100-m-apart-point-midway-point-b-800-10-2-m-q1-600-10-q7775023

MY WRONG ANSWERS FOR PART C, ( PLEASE DONT REPEAT THEM ) :

1.22 * 10 ^-7

- 1.22 * 10^-7

1.22

1.24

Explanation / Answer

as your answer for part a and c are not visible.I am solving full question for you

Electric potential at A is

VA = V1A + V2A

=kq1/r1a+kq2/r2a

=9*10^9((2.15*10^-9/0.050m)+(-6.5*10^-9/0.050))

=9*10^9(43*10^-9-130*10^-9)

Va=-783J/C

Vb=V1b+V2b

=kq1/r1b+kq2/r2b

=9*10^9(2.15*10^-9/0.080m)+(-6.5*10^-9/0.060))

=9*10^9(26.8*10^-9-108.3*10^-9)

Vb=-733.5J/C

c) now work done is

UBA = qVBA = q(VA VB)

=2.15*10^-9(-783+733.5)

=-106.4*10^-9

=-106nJ--answer

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