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December 21, 2014

December 21, 2014

Posted by **susane** on Monday, March 12, 2007 at 6:22pm.

a) 0.040 C

b) 12.5 C

c) 20C

D) cannot be computed unless the path is give

e) none of these

My solution:

I used the equation V=W/q inorder to solve for the charge. I plugged in the numbers to get 40= (500)/X. The answer I got was .08. According to the book the answer is 12.5 C

The potential difference between two points is 100V. If 2 C is transported from one of these points to the other, the magnitude of work done is:

a) 200J

b) 100 J

c) 50J

d) 100 J

e) 2 J

The solution I figured would use the same equation because involves potential difference and charge. When I calculated the answer using V=W/q I came up with the answer I came up with A 200J.

Am I handling these problems right because in the second question my answer was right but not in the first one.

Since 40 = 500/X, then X = 500/40 = 12.5 C

Your math is unique.

QV= energy

Q= 500/40 Coulombs That is not .08C

The second is correct. QV=W

- Physics -
**Magoo**, Sunday, April 29, 2012 at 10:23pmIn fact q= W/ V

This is q= 500J/40V

so, q= 12.5J/V or 12.5 Coulombs (C)

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