For a single, isolated point charge carrying a charge of q=1.92×10?11 C, one equ

Question

For a single, isolated point charge carrying a charge of q=1.92×10?11 C, one equipotential surface consists of a sphere of radius r1=0.0154 m centered on the point charge as shown in the figure. What is the potential on this surface?

Now consider an additional equipotential surface that is separated by 4.89 V from the previously mentioned surface. How far from the point charge should this surface be? This surface must also meet the condition of being farther from the point charge than the original equipotential surface. distance from point charge: m

Explanation / Answer

For a point charge, potential at a distance r is

V=kQ/r

V=9*109*1.92*10-11/0.0154

V=11.22V

Potential difference is,

V=11.22-4.89

V=6.33V

So now,

6.33=9*109*1.92*10-11/r

r=0.0273 m

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