Charge Q=4 micro-coulombs is distributed uniformly over the volume of an insulating sphere of radius R=.05 meters. What is the potential difference between the center of the sphere and the surface of the sphere?
How do I approach this problem? I tried integrating dV=-Edr from r=infinity (V=0) to r=.05 but didn't get the right answer. Should I be integrating with respect to the volume?... the text book said the answer was 360 kV.
To approach this problem, you can use the concept of potential difference and the symmetric nature of the charge distribution on the insulating sphere. Here's the step-by-step explanation on how to solve it:
1. Start by considering a small elemental charge dQ within the sphere. This elemental charge can be assumed to be located at a distance r from the center of the sphere.
2. The electrical potential due to this elemental charge at a point on the surface of the sphere can be calculated using the formula for electrical potential of a point charge V = k * Q / r, where k is the electrostatic constant (8.99 x 10^9 Nm²/C²) and Q is the charge of the elemental charge.
3. Considering the elemental charge as a point charge at the sphere's surface, the distance between the center of the sphere and this elemental charge will be R, the radius of the sphere.
4. Now, integrate the electrical potential due to all the elemental charges on the sphere's surface to find the total potential at the surface. Since the charge distribution is uniform, you can treat the integration as summing the potential contributions from each elemental charge.
5. The integral to be evaluated is V = ∫(k * Q / R) * dQ, where dQ represents a small elemental charge on the sphere's surface ranging from 0 to Q.
6. Substituting the given values, Q = 4 micro-coulombs, R = 0.05 meters, and k = 8.99 x 10^9 Nm²/C², the integral becomes V = ∫(8.99 x 10^9 * (4 x 10^-6) / 0.05) * dQ.
7. Now, you can integrate the above expression with respect to the charge. The limits of integration will be from 0 to Q = 4 micro-coulombs.
8. Simplify the expression, and the result will be the potential difference between the center and the surface of the sphere.
By following these steps, you should be able to calculate the correct potential difference.