Answer: Potential Difference Va-Vb for 4.3 nC Particle in 225 N/C Field

AI Thread Summary
The discussion revolves around calculating the potential difference Va - Vb for a particle with a charge of +4.3 nC in a uniform electric field of 225 N/C. Key points include the necessity of knowing the distance between points a and b and their alignment with the electric field to accurately determine the potential difference. The electric force on the particle was calculated to be 0.968 N, and the work done was found to be 0.242 J. It was emphasized that potential difference is measured in volts, which equate to joules per coulomb. The distance between points a and b was confirmed to be 0.25 m, leading to the successful calculation of the potential difference.
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Homework Statement



a uniform electric field has magnitude E = 225 N/C and is directed to the right. A particle with charge +4.3 nC moves along the straight line from a to b.
What is the potential difference Va - Vb between points a and b?

Homework Equations



Potential dif=Va-Vb

The Attempt at a Solution



I found the electric force and the work done on the particle. electric force is .968 and the work done is .242
 
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Four comments:

1. We would need to know the distance between a & b, and whether they lie "along" the field line or at some angle to the field, in order to answer the question.

2. You didn't put units with your answers.

3. Are you aware how much a nanoCoulomb is?

4. Electric potential difference can be determined from the work down and the charge.
Hint: units of potential difference are Volts, which are equivalent to Joules per Coulomb.
 
the distance is .25 m.
 
I figured it out. Thanks a lot!
 

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