SolPr1

Question1

We consider the cylinder as two cylindrical capacitors in parallel. The two “negative plates” are the (connected) halves of the inner cylinder (half of which is in contact with liquid, and half of which is in contact with vapor). The two “positive plates” are the (connected) halves of the outer cylinder (half of which is in contact with liquid, and half of which is in contact with vapor).

For the cylindrical capacitor

$V=V_{b}-V_{a}=-\int _{a}^{b}{\vec {E}}.d{\vec {l}}=-{\frac {Q}{2\pi \epsilon _{0}l}}\int _{R_{a}}^{R_{b}}{\frac {dR}{R}}={\frac {Q}{2\pi \epsilon _{0}l}}\ln {\frac {R_{a}}{R_{b}}}$

$C={\frac {Q}{V}}={\frac {2\pi \epsilon _{0}l}{\ln(R_{a}/R_{b})}}$

(a)

$C=C_{\textrm {liq}}+C_{\textrm {V}}={\frac {2\pi \epsilon _{0}K_{\textrm {liq}}h}{\ln(R_{a}/R_{b})}}+{\frac {2\pi \epsilon _{0}K_{\textrm {V}}(l-h)}{\ln(R_{a}/R_{b})}}={\frac {2\pi \epsilon _{0}l}{\ln(R_{a}/R_{b})}}\left[(K_{\textrm {liq}}-K_{\textrm {V}}){\frac {h}{l}}+K_{\textrm {V}}\right]=C$

${\frac {h}{l}}={\frac {1}{(K_{\textrm {liq}}-K_{\textrm {V}})}}\left[{\frac {C\ln(R_{a}/R_{b})}{2\pi \epsilon _{0}l}}-K_{\textrm {V}}\right]$

(b)

For the full tank, ${\frac {h}{l}}=1$, for the empty tank ${\frac {h}{l}}=0$

Full=$1.5\times 10^{-9}$F Empty=$1.1\times 10^{-9}$F

Question 2

The kinetic energy of the electrons (provided by the UV light) is converted completely to potential energy at the plate since they are stopped. Use energy conservation to find the emitted speed, taking the 0 of PE to be at the surface of the barium.

${\textrm {KE}}_{\textrm {initial}}={\textrm {PE}}_{\textrm {final}}\rightarrow qV={\frac {1}{2}}mv_{x}^{2}$

$v=1.03\times 10^{6}$m/s