Respuesta :
Explanation:
(1). Formula to calculate the potential difference is as follows.
[tex]\Delta V = -\int E dr[/tex]
= [tex]-\int \frac{kq}{r^{2}} dr[/tex]
= [tex]\frac{kq}{r_{f}} - \frac{kq}{r_{i}}[/tex]
= [tex]\frac{kq(r_{f} - r_{i})}{r_{f}r_{i}}[/tex]
= [tex]\frac{9 \times 10^{9} \times 3.30 \times 10^{-9}(0.1 - 0.015)}{0.1 \times 0.015}[/tex]
= 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 = [tex]\frac{1}{2}QV[/tex]
= [tex]\frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V[/tex]
= [tex]6.39 \times 10^{-8} J[/tex]
Thus, the electric-field energy stored in the capacitor is [tex]6.39 \times 10^{-8} J[/tex].
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 .
[tex]\Delta V = -\int E.dr[/tex]
[tex]\Delta V = -\int_{r'}^{r}\frac{Kq}{r}dr[/tex]
[tex]\Delta V = \frac{kq}{r'}- \frac{kq}{r}[/tex]
[tex]\Delta V = 9\times 10^{9}\times 3.3 \times 10^{-9}\times \left ( \frac{1}{0.015}-\frac{1}{0.1} \right )[/tex]
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
