Potential of a uniformyl charged sphere

[ehabmozart]

Homework Statement

The question asks to find the potential due to a uniformly charged sphere at r<R where R is the radius.

Homework Equations

Va-Vb= ∫ E dr from a → b

The Attempt at a Solution

My attempt was like this

Va is to be a the point of interest r.. Vb to be R... We know Vb to be KQ/R... Now E inside the sphere is KQr/R^3 .. After integration with respect to r we get (R^2-r^2)*Kq/2R^3 ... Thus the net potential at Va is Vb + the integral which is KQ/R + (R^2-r^2)*Kq/2R^3 ... However, this isn't the answer given.. Where was my mistake??

 

[mfb]

How does it differ from the answer given?
Note that it is possible to simplify your expression.

 

[ehabmozart]

How does it differ from the answer given?
Note that it is possible to simplify your expression.

I am not able to simplify.. Can you help me?

 

[mfb]

Just use (a-b)/c = a/c - b/c for (R^2-r^2)*Kq/(2R^3).

 

[Nickie]

Your attempt at solving this problem is correct, however, there may be a few small errors in your calculations. First, the electric field inside a uniformly charged sphere is given by E = (KQr)/R^3, not KQr/R^3. This may have resulted in a slight error in your integration. Additionally, when integrating from a point r to R, the limits of integration should be r and R, not 0 and R. Finally, when calculating the potential at point Va, you should subtract the potential at point Vb, not add it. This will result in the final answer of Va = (R^2 - r^2)KQ/2R^3. Keep in mind that this is only valid for points inside the sphere, as the potential outside the sphere is simply V = KQ/r. Overall, your approach and understanding of the problem is correct, just be sure to double check your calculations and take into account the limits of integration and the direction of the potential difference.