Answer:
The question is incomplete, below is the complete question
"Two equipotential surfaces surround a +1.38*10^-8C point charge. How far is the 179-V surface from the 67.5-V surface? "
Answer:
1.15m
Explanation:
From the formula connecting the charge, potential and the distance i.e
[tex]V=\frac{kq}{r}[/tex]
where q=charge, k=constant and r=distance between the points
the data given in the question are
[tex]q=1.38*10^{-8}c,\\V_{1}=179v\\V_{2}=67.5v\\r_{1}-r_{2}=??[/tex]
We can express the distance from the charge to the equipotential surface as
[tex]r_{1}=\frac{kq}{v_{1}}\\[/tex]
for the second charge we have
[tex]r_{2}=\frac{kq}{v_{2}}\\[/tex]
The distance between the two surface is
[tex]r_{2}-r_{1}=\frac{kq}{v_{2}}-\frac{kq}{v_{1}}\\r_{2}-r_{1}=kq(\frac{1}{v_{2}}-\frac{1}{v_{1}})\\r_{2}-r_{1}=(9*10^{9}*1.38*10^{-8})(\frac{1}{67.5}-\frac{1}{179})\\r_{2}-r_{1}=124.2(0.0148-0.00558)\\r_{2}-r_{1}=1.15m[/tex]
Hence the distance is 1.15m
