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elena-s [515]
3 years ago
11

Two equipotential surfaces surround a +2.90 x 10-8-C point charge. How far is the 220-V surface from the 85.0-V surface?

Physics
1 answer:
RUDIKE [14]3 years ago
4 0
The potential difference isV=220−85=135.
Now the potential V at distance d from a point charge Q isV=Q/4πϵ0d
If you solve this for d and plug in the potential difference you will get the answer.
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The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

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