Question

A spherical capacitor is formed from two concentric sphericalconducting shells separated by vacuum. The inner sphere has radius10.0 centimeters, and the separation between the spheres is 1.50 centimeters. The magnitude of the charge on each sphere is 3.30 nanocoulombs.1)What is the magnitude of the potential difference DeltaV between the two spheres?2)What is the electric-field energy stored in the capacitor?

Answer 1

Explanation:

(1).  Formula to calculate the potential difference is as follows.

       $\Delta V = -\int E dr$

                  = $-\int \frac{kq}{r^{2}} dr$

                 = $\frac{kq}{r_{f}} - \frac{kq}{r_{i}}$

                 = $\frac{kq(r_{f} - r_{i})}{r_{f}r_{i}}$

                 = $\frac{9 \times 10^{9} \times 3.30 \times 10^{-9}(0.1 - 0.015)}{0.1 \times 0.015}$

                = 38.7 volts

Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.

(2).  Now, formula to calculate the energy stored in the capacitor is as follows.

           E = $\frac{1}{2}QV$

              = $\frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V$

              = $6.39 \times 10^{-8} J$

Thus, the electric-field energy stored in the capacitor is $6.39 \times 10^{-8} J$.

Answer 2

Answer:

Explanation:

r = 10 cm = 0.1 m

r' = 1.5 cm = 0.015 m

Q = 3.3 x 10^-9 C

1.

Let the potential difference is ΔV .

$\Delta V = -\int E.dr$

$\Delta V = -\int_{r'}^{r}\frac{Kq}{r}dr$

$\Delta V = \frac{kq}{r'}- \frac{kq}{r}$

$\Delta V = 9\times 10^{9}\times 3.3 \times 10^{-9}\times \left ( \frac{1}{0.015}-\frac{1}{0.1} \right )$

ΔV = 1683.1 V

2.

Energy stored, E = 0.5 x Q x V

E = 0.5 x 3.3 x 10^-9 x 1683.1

E = 2.78 x 10^-6 J