Question

The two charges in the figure below are separated by d = 4.00 cm. (Let q1 = -20 nC and q2 = 26.5 nC.)

figure link:
http://assets.openstudy.com/updates/attachments/4f28ac29e4b049df4e9d71a4-chansharp-1328065664864-25p012alt.gif

Answer 1

what's the question?

Answer 2

(a) Find the electric potential at point A.
(b) Find the electric potential at point B, which is halfway between the charges.

im sorry, haha... I kind of get what their asking but I just cant get the answer right.

Answer 3

Im using V=((Ke)(q1))/(r) + ((Ke)(q2))/(r)

Answer 4

kq/r for each charge, where r is the distance from the point to the charge (ie r=d at point A)

Answer 5

yes, that's right.

Answer 6

thanks for your help... I just need alittle more help here... Im getting 2924.65 for an answer but it says its wrong!?

Answer 7

Is there a chance that you can work it out?

Answer 8

seems high.

Answer 9

2924 V
29 kV

Answer 10

k( ( -20nC/.04) + (26.5nC/.04) )

Answer 11

okkk thank you very much

Answer 12

where k ~ 9E9

Answer 13

thanks alot

Answer 14

get the right answer?

Answer 15

Im working on it... for A I got 1462 V (Tried this question too many times so its already closed, so I dont know if its right)
and Im working for B

Answer 16

distance on B is going to be .02 right? instead of .04

Answer 17

yes:)

Answer 18

I got 2924 V for B

Answer 19

is that right?>

Answer 20

GOT IT! :) THanks!

Answer 21

sure!

Answer 22

coulombs law Force F=k q1q21/r2

Answer 23

@masumanwar wrong.