Question

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?

Answer 1

Electric potential of the second conductive metal sphere will be
`(1)/(5)` times than first conductive metal

Step-by-step 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

`V=(1)/(4\pi \epsilon _0)(Q)/(R)`

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
`(1)/(5)` times than first conductive metal