C-2-0-L-ln-R-2-R-1-prove-

Question Number 60219 by ANTARES VY last updated on 19/May/19

C = \frac{2 π ε ε_{0} L}{\ln (\frac{R_{2}}{R_{1}})} .

prove .

Commented by ANTARES VY last updated on 19/May/19

Answered by ajfour last updated on 19/May/19

C = \frac{q}{V} = \frac{q}{\int_{R_{1}}^{R_{2}} E . dr}

from {Gauss}^{'} law E = \frac{q / L}{2 π ϵ ϵ_{0} r}

so C = \frac{q}{V} = \frac{q}{\int_{R_{1}}^{R_{2}} \frac{(q / L)}{2 π ϵ ϵ_{0} r} dr}

\Rightarrow C = \frac{2 π ϵ ϵ_{0} L}{\int_{R_{1}}^{R_{2}} \frac{dr}{r}} = \frac{2 π ϵ ϵ_{0} L}{\ln (\frac{R_{2}}{R_{1}})} .

Commented by ANTARES VY last updated on 19/May/19

?

Commented by ajfour last updated on 19/May/19

Calculate Electric field from

{Gauss}^{'} law, rest i have explained

how to use that result .

Commented by ANTARES VY last updated on 19/May/19

E = \frac{\frac{q}{L}}{2 π ε ε_{0} r} \frac{q}{L} = ? . \frac{1}{2 π ε ε_{0} r} = \frac{2 k}{ε r} = \frac{2}{C}

C = \frac{ε R}{k} .

Commented by ajfour last updated on 19/May/19

Commented by ajfour last updated on 19/May/19

According to {Gauss}^{'} law

flux of field through a Gaussian

surface = \frac{1}{ϵ_{0}} (charge within Gaussian surface)

assuming charge q and - q on capacitor

plates

ϵ E (2 π rL) = \frac{1}{ϵ_{0}} (q)

\Rightarrow E = \frac{q / L}{2 π ϵ ϵ_{0} r} .

Commented by ANTARES VY last updated on 19/May/19