Question

electric potential due to charged sphere at interior and exterior point

Answer 1

We know what the electric field and potential from a point charge look like:

E=kQr2    and    V=kQr

Consider a charged sphere with a symmetrical distribution of charge. Gauss' Law tells us that the electric field outside the sphere is the same as that from a point charge. This implies that outside the sphere the potential also looks like the potential from a point charge.

What about inside the sphere? If the sphere is a conductor we know the field inside the sphere is zero. What about the potential?

Moving from a point on the surface of the sphere to a point inside, the potential changes by an amount:

ΔV = -∫ E • ds

Because E = 0, we can only conclude that ΔV is also zero, so V is constant and equal to the value of the potential at the outer surface of the sphere.

Potential near an Insulating Sphere

Now consider a solid insulating sphere of radius R with charge uniformly distributed throughout its volume. Once again, outside the sphere both the electric field and the electric potential are identical to the field and potential from a point charge.

What happens inside the sphere? Now the potential is not constant because there is a field inside the sphere. Using Gauss' Law we showed that the field inside a uniformly charged insulator is:

E=k Q rR3

Use this to calculate the potential inside the sphere. Starting from some point a distance r from the center and moving out to the edge of the sphere, the potential changes by an amount:

ΔV = V(R) - V(r)  =–∫  R    E • dsr  =  –∫  R    E drr  =  –kQR3∫  R    r drr

Integrating gives:

V(R) – V(r)=– kQ2R3(R2 – r2)

V(R) – V(r)=– kQ2R( 1 –r2R2)

V(R) is simply kQ/R, which can be written as 2kQ/2R, so:

V(r)=2kQ2R+kQ2R( 1 –r2R2)

Therefore, for r < R,

V(r)=kQ2R( 3 –r2R2)

Hope it helps.

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