Poll Results
No votes. Be the first one to vote.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
No votes. Be the first one to vote.
c
Explanation: Vab = (Q/4πεo)(1/rA) + (1/rB), where rA and rB are position vectors rA =
1m and rB = 4m. Thus Vab = 2.7 volts.
The electric potential (V) due to a point charge (q) at a distance (r) in a vacuum is given by the formula:
[ V = frac{1}{4piepsilon_0} cdot frac{q}{r} ]
where (epsilon_0) is the vacuum permittivity constant, approximately equal to (8.85 times 10^{-12} , text{C}^2/text{N}cdottext{m}^2).
The potential difference (Delta V) between two points due to a point charge is the difference in the electric potentials at those two points, given by:
[ Delta V = V_2 – V_1 ]
Considering the point charge of (0.4 text{nC}) (or (0.4 times 10^{-9} , text{C})) located at ((2, 3, 3)) meters, we want to find the potential difference between points ((2, 3, 3)) meters and ((-2, 3, 3)) meters.
### Calculation
1. Distance of the first point from the charge:
The first point is the location of the charge itself, so (r_1 = 0).
– For practical purposes, the potential at the location of a point charge is infinite, but since we’re calculating a