Consider two isolated conductive metal spheres. Each carries the same amount of excess charge +Q, but one has a radius that is five times greater than the other.

How does the electrostatic potential of the two spheres compare?

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aristeus aristeus · Answer 1 · 2019-12-10T09:10:30+01:00

Answer:

Electric potential of the second conductive metal sphere will be [tex]\frac{1}{5}[/tex] times than first conductive metal

Explanation:

We have given that there are there are two isolated conductive metal surfaces each have charge +Q

And one has radius 5 times greater than second

Now we know that electric potential due to conducting sphere

[tex]V=\frac{1}{4\pi \epsilon _0}\frac{Q}{R}[/tex]

As from the relation we can see that electric potential is inversely proportional to the radius of the sphere

So electric potential of the second conductive metal sphere will be [tex]\frac{1}{5}[/tex] times than first conductive metal