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Physics Question on electrostatic potential and capacitance

A cylindrical capacitor has two co-axial cylinders of length 20 cm20 \,cm and radii 1.5 cm1.5\, cm and 1.6 cm1.6 \,cm. The outer cylinder is earthed and inner cylinder is given a charge of 4 μC4\, \mu C. The capacitance of the system is (neglect end effects)

A

2.8×10−8 F2.8 \times 10^{-8} \, F

B

4.2×10−14 F4.2 \times 10^{-14} \, F

C

1.7×10−10 F1.7 \times 10^{-10} \, F

D

3.4×10−12 F3.4 \times 10^{-12} \, F

Answer

1.7×10−10 F1.7 \times 10^{-10} \, F

Explanation

Solution

Here, length L=20 cm=20×10−2 mL = 20\, cm = 20 \times10^{-2} \, m inner radius ri=1.5 cm=1.5×10−2 mr_{i}=1.5\,cm=1.5\times10^{-2}\, m outer radius r0=1.6 cm=1.6×10−2 m r_{0}=1.6 \,cm=1.6\times10^{-2}\,m charge q−4 μC=4×10−6 Cq - 4 \, \mu C=4\times10^{-6}\, C Capacitance, C=2πε0Lloge(r0ri)=2πε0L2.3 log10(r0ri)C=\frac{2\pi\varepsilon_{0} L}{log_{e}\left(\frac{r_{0}}{r_{i}}\right)}=\frac{2\pi\varepsilon_{0} L}{2.3 \, log_{10} \left(\frac{r_{0}}{r_{i}}\right)} =2π×8.85×10−12×20×10−22.3log10(1.6×10−21.5×10−2)=\frac{2\pi\times8.85\times10^{-12}\times20\times10^{-2}}{2.3 log_{10} \left(\frac{1.6\times10^{-2}}{1.5\times10^{-2}}\right)}\quad =1.7×10−10 F=1.7\times10^{-10}\, F