Question

Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of dielectric oil (of mass density r, dielectric constant K). The inner one is maintained at potential V and the outer one is grounded. To what equilibrium height (h) does the oil rise in the space between the tubes ? [Assume this height (h) as a equilibrium height]

• A. $\frac{\varepsilon_{0}2V^{2}(K - 1)}{g\rho(b^{2} - a^{2})\log\frac{b}{a}}$
• B. $\frac{\varepsilon_{0}V^{2}(K - 1)}{\rho(b^{2} - a^{2})g\log\frac{b}{a}}$
• C. $\frac{4\varepsilon_{0}V^{2}(K - 1)}{g\rho(b^{2} - a^{2})\log\frac{b}{a}}$
• D. $\frac{6\varepsilon_{0}V^{2}(K - 1)}{\rho(b^{2} - a^{2})g\log\frac{b}{a}}$

Answer 1

Answer: (B)

$\frac{\varepsilon_{0}V^{2}(K - 1)}{\rho(b^{2} - a^{2})g\log\frac{b}{a}}$

Answer 2

Gravitational force = mg

= rp (b² – a²)gh

Net upward force F = $\frac { \mathrm { dU } } { \mathrm { dh } }$

=

where h is the height of liquid.

Now we calculate C as a function of h

C_eq = C_Air + C_K

= $\frac { 2 \pi \varepsilon _ { 0 } ( \mathrm {~L} - \mathrm { h } ) } { \ln \frac { \mathrm { b } } { \mathrm { a } } }$ +

C = $\frac { 2 \pi \varepsilon _ { 0 } \{ \mathrm {~L} + ( \mathrm { K } - 1 ) \mathrm { h } \} } { \ln \frac { \mathrm { b } } { \mathrm { a } } }$

F = =$\frac { 1 } { 2 } \mathrm {~V} ^ { 2 }$ื

\ F = mg

$\frac { \frac { 1 } { 2 } \mathrm {~V} ^ { 2 } \times 2 \pi \varepsilon _ { 0 } ( \mathrm {~K} - 1 ) } { \log _ { \mathrm { e } } \frac { \mathrm { b } } { \mathrm { a } } }$= rp(b² – a²) gh

h = $\frac { \pi \varepsilon _ { 0 } \mathrm {~V} ^ { 2 } ( \mathrm {~K} - 1 ) } { \rho \pi \left( \mathrm { b } ^ { 2 } - \mathrm { a } ^ { 2 } \right) \mathrm { g } \log _ { \mathrm { e } } \frac { \mathrm { b } } { \mathrm { a } } }$

= $\frac { \varepsilon _ { 0 } V ^ { 2 } ( K - 1 ) } { \rho \left( b ^ { 2 } - a ^ { 2 } \right) g \log _ { e } \frac { b } { a } }$