211k views
4 votes
A sphere of radius R has a charge Q uniformly distributed throughout its volume. Show that for rV(r)= kQ(3R²−r²)/2R³

(Hint: The electric field within a uniformly charged sphere is E=kQr/R³. Evaluate V(r)−V(R).)

1 Answer

2 votes

The potential V(r) inside a uniformly charged sphere is found by integrating the known electric field E=kQr/R³ with respect to r, yielding V(r) = kQ(3R² - r²)/2R³, indicating the potential's dependence on the distance from the center.

The question relates to the electric field and potential within and around a uniformly charged sphere. To derive the potential V(r) inside the sphere, one can start from the known electric field E within a uniformly charged sphere, given by E=kQr/R³, where k is Coulomb's constant, Q is the total charge, R is the sphere's radius, and r is the distance from the center of the sphere to the point of interest. Since the electric potential V is defined as the negative integral of the electric field E with respect to distance r, we can express V(r) - V(R) as the integral of E from R to r.

For the electric potential difference, we have ∆V = V(r) - V(R) = - ∫ E · dr, which after integration and applying the limits from R to r, gives us V(r) = kQ(3R² - r²)/2R³. This result indicates that at any point inside the sphere, the potential depends on the square of r, decreasing as we move towards the center. Such a calculation is important for understanding electric fields in materials with spherical symmetry, like charged particles or celestial bodies.

User Statmaster
by
7.4k points