# Capacitance of spherical capacitor when inner surface of outer sphere is earthed

Let there be two concentric shells in which the outer sphere contains charge Q1 and inner sphere contains charge Q.Capacitance of spherical capacitor when the inner surface of the outer sphere is earthed is: $$C={4\pi \varepsilon_{0}b}.$$

But the charge present in the outer surface of the inner sphere(Q) has a potential (V). Hence there is a Potential Difference between the outer surface of the inner sphere and the inner surface of the outer-sphere. Potential difference :$$V=\frac{Q}{4\pi \varepsilon_{0}a}$$ So the capacitance is $$C1=\frac{Q}{V} = {4\pi \varepsilon_{0}a}$$

And total capacitance: $$C={4\pi \varepsilon_{0}a}+{4\pi \varepsilon_{0}b}$$ where does the derivation is go wrong?

• From where did you get that potential difference? – Ofek Gillon Apr 13 at 18:10
• the potential of the inner sphere minus the potential of inner surface of outer-sphere = V-0 =V – Sankara Narayan Apr 14 at 6:06
• The voltage on the outer sphere isn't what you wrote – Ofek Gillon Apr 14 at 9:15
• earthed surface must have potential equal to zero ? – Sankara Narayan Apr 15 at 20:43

Grounding the inner surface of the outer shell of a concentric spherical capacitor sets the voltage at that surface to zero. Since we have the reference voltage defined as $$V_{R\rightarrow\infty}=0$$ this means there is no electric field outside of the system of concentric shells. What this implies is that if you place charge Q on the inner sphere, opposite and equal charge -Q will flow onto the inner surface of the outer sphere such that it negates the field outside of the system. It turns out that this is the same picture as the one we get in the link above, the only difference being there is no electric field once you pass outside of the inner surface of the outer shell. Thus the capacitance should come out the same:$$C= \frac{4 \pi \epsilon_0}{\frac{1}{a}-\frac{1}{b}}.$$ It's important to understand that the voltage or charge placed on the outer shell has no bearing on the capacitance, since by Gauss' Law the field is determined solely by the charge contained within the Gaussian surface in question.