A cylindrical capacitor has two co-axial cylinders of length $15\; cm$ and radii $1.5 \;cm$ and $1.4\; cm .$ The outer cylinder is earthed and the inner cylinder is given a charge of $3.5\; \mu \,C .$ Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
A cylindrical capacitor has two co-axial cylinders of length $15\; cm$ and radii $1.5 \;cm$ and $1.4\; cm .$ The outer cylinder is earthed and the inner cylinder is given a charge of $3.5\; \mu \,C .$ Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
Length of a co-axial cylinder, $l=15 \,cm =0.15\, m$
Radius of outer cylinder, $r_{1}=1.5 \,cm =0.015\, m$
Radius of inner cylinder, $r_{2}=1.4 \,cm =0.014 \,m$
Charge on the inner cylinder, $q=3.5\, \mu \,C=3.5 \times 10^{-6} \,C$
Capacitance of a co-axial cylinder of radii $r_{1}$ and $r_{2}$ is given by the relation
$C=\frac{2 \pi \epsilon_{0} l}{\log _{e r_{2}}^{r_{1}}}$
Where, $\varepsilon_{0}=$ Permittivity of free space $=8.85 \times 10^{-12}\, N ^{-1} \,m ^{-2} \,C ^{2}$
$\therefore C =\frac{2 \pi \times 8.85 \times 10^{-12} \times 0.15}{2.3026 \log _{10}\left(\frac{0.15}{0.14}\right)}$
$=\frac{2 \pi \times 8.85 \times 10^{-12} \times 0.15}{2.3026 \times 0.0299}$$=1.2 \times 10^{-10} \,F$
Potential difference of the inner cylinder is given by,
$V=\frac{q}{C}$
$=\frac{3.5 \times 10^{-6}}{1.2 \times 10^{-10}}=2.92 \times 10^{4}\, V$
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- [KVPY 2019]
Answer the following:
$(a)$ The top of the atmosphere is at about $400\; kV$ with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about $100\; Vm ^{-1} .$ Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
$(b)$ A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area $1\; m ^{2} .$ Will he get an electric shock if he touches the metal sheet next morning?
$(c)$ The discharging current in the atmosphere due to the small conductivity of air is known to be $1800 \;A$ on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
$(d)$ What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (The earth has an electric field of about $100\; Vm ^{-1}$ at its surface in the downward direction, corresponding to a surface charge density $=-10^{-9} \;C \,m ^{-2} .$ Due to the slight conductivity of the atmosphere up to about $50\; km$ (beyond which it is good conductor), about $+1800 \;C$ is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Answer the following:
$(a)$ The top of the atmosphere is at about $400\; kV$ with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about $100\; Vm ^{-1} .$ Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
$(b)$ A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area $1\; m ^{2} .$ Will he get an electric shock if he touches the metal sheet next morning?
$(c)$ The discharging current in the atmosphere due to the small conductivity of air is known to be $1800 \;A$ on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
$(d)$ What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (The earth has an electric field of about $100\; Vm ^{-1}$ at its surface in the downward direction, corresponding to a surface charge density $=-10^{-9} \;C \,m ^{-2} .$ Due to the slight conductivity of the atmosphere up to about $50\; km$ (beyond which it is good conductor), about $+1800 \;C$ is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)