Question

NO QUESTIONS DURING THE EXAM! e capacitor with square plates of area A 4.00 x 10-2 m2 and h a dielectric having 3. It is fully charged to a voltage of 100 V and the (35 points) A parallel plate separation with charging battery is then disconnected. Take o a) the capacitance C b) the charge in the capacitor Q. c) the magnitude of the electric field E between the plates. d) After the battery has been disconnect 8.85 10 CN m. Find ed the separation between the plates is increased to d ential V -8.00 mm .as shown in the second figure. Calculate the value of the capacitance C and the po V-100 V K-3 d' ?' K-3

Answer 1

(a)

$C=\frac {kA\varepsilon_0}{d}=\frac {3\times 4\times 10^{-2}\times 8.854\times 10^{-12}}{4\times 10^{-3}}\: F$

(b)

$Q=CV=(0.267\times 100) \: nC = 26.7\: nC$

(c)

$E=\frac {V}{d}=\frac {100}{0.004}\: V/m = 2.5\times 10^5\: V/m$

(d)

$C'=\frac {\frac {kA\varepsilon_0}{d/2}\cdot \frac {A\varepsilon_0}{d/2}}{\frac {kA\varepsilon_0}{d/2}+\frac {A\varepsilon_0}{d/2}}$

$C'=\frac {k}{k+1}\frac {A\varepsilon_0}{d}$

$C'=\frac {3}{3+1}\times \frac { 4\times 10^{-2}\times 8.854\times 10^{-12}}{4\times 10^{-3}}$

$Q=Q'$

$CV=C'V'$

$V'=\frac {C}{C'}V=\frac {0.267}{0.200}\times 100\: V$