Electric Potential questions and work done

AI Thread Summary
The discussion focuses on calculating electric potential energy at two points, A and B, from a +1.0μC charge, using the formula V=kQ/r. The electric potential at A is calculated as approximately 2.997 x 10^9 V, while at B it is about 1.499 x 10^9 V. The work done moving a +0.2μC charge between A and B is calculated to be approximately 0.00029966 J. Participants clarify that the units for electric potential are volts and for work done are joules, correcting earlier misinterpretations. The conversation emphasizes understanding the relationship between electric potential and work in electric fields.
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Homework Statement




+1.0uC
*-------3.0m---------*-----------3.0m---------*
A B
a) calculate the electric potential energy at A, 3.0m away from a point charge of +1.0uC.
b) calculate the electric potential energy at B, 6.0m away from a point charge of +1.0uC.
c) how much work would be done moving a +0.2uC charge between A and B?
d) which way would you be moving this smal charge if you were doing work against the field?

Homework Equations



V=kQ/r
W-qV


The Attempt at a Solution



a)
V=kQ/r
V= 8.99 x 10^9 x 1.0 / 3 = 2996666667N

b)
V=kQ/r
V=8.99 x 10^9 x 1.0 / 6 = 1498333333N

c)
W=qV
W= 8.99 x `10^9 x .2
W = 1798000000

d)
to the right.

I'm not sure if I'm on the right track, can anyone check and help me?
 
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Firstly, the charge is 1μC (10-6 C) and you have used a value of 1
The formula is correct.
The answer is electric potential and this is measure in volts. You have stated N which means Newtons. (Force)

The formula is correct for the next part but for W=qV
V is the potential difference in volts between the two points, and q the charge you move.
You will have a value for V when you have got the answers to parts a and b.
 
HMm.. thank you.. is this better...

V=kQ/r
8.99 x 10^9 x 1.0x 10^-6/ 3

and

8.99 x 10^9 x 1.0 x 10^-6/6

for W=qV
= .2 x 10^-6 x 1498.3
=0.00029966N
 
Yes that's the correct answer for the 3rd part but the unit is joule (J) because you have calculated the work done.
In this case the field does the work on the charge.
 
Ok thankyou brilliant. I'm getting basics confused.
 
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