Answer:
[tex]C=\frac{ab}{k(b-a)}[/tex]
Explanation:
We can assume this problem as two concentric spherical metals with opposite charges.
We have also to take into account the formulas for the electric field and the capacitance. Hence we have
[tex]C=\frac{Q}{V}\\\\E=k\frac{Q}{r^2}\\[/tex]
Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating
[tex]dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}][/tex]
Hence, the capacitance is
[tex]C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}[/tex]
but R1=a and R2=b
[tex]C=\frac{ab}{k(b-a)}[/tex]
HOPE THIS HELPS!!
Answer:
The capacitance of a spherical capacitor can be give by [tex]C = \frac{ab}{k(b-a)}[/tex]
Explanation:
Let Q = Magnitude of the charge on the spherical conductor
a = Radius of the smaller sphere
b = Radius of the larger sphere
According to Gauss' law, [tex]E [4\pi r^{2} ] = \frac{Q}{\epsilon_{0} }[/tex]
[tex]E(r) = \frac{Q}{4\pi \epsilon_{0} r^{2} }[/tex]
The electric potential between the spherical conductor is given by :
[tex]V = \int\limits^a_b {E(r)} \, dr[/tex]
[tex]V = \frac{Q}{4\pi \epsilon_{0} } \int\limits^a_b {\frac{1}{r^{2} } } \, dr[/tex]
[tex]V = \frac{Q}{4\pi \epsilon_{0} }[\frac{1}{a} - \frac{1}{b} ]\\V = \frac{Q}{4\pi \epsilon_{0} }[\frac{b-a}{ab} ][/tex]
But [tex]k = \frac{Q}{4\pi \epsilon_{0} }[/tex]
[tex]V = kQ[\frac{b-a}{ab} ][/tex]
The capacitance of the spherical capacitor can be given by [tex]C = \frac{Q}{V}[/tex]
[tex]C = \frac{Q}{ kQ[\frac{b-a}{ab} ]} \\C = \frac{ab}{k(b-a)}[/tex]
