Question

a) A capacitor is made of two circular plates of metal with radius r separated by distance d. The gap between them is vacuum. What is the capacitance of the capacitor? (5 pts) b) A long wire is hooked up to the capacitor as shown in figure 1 below. A current runs through the wire to charge the capacitor. At what rate is the electric field between the capacitor plates changing? (5 pts) Figure 1: Charging Capacitor

Answer 1

a] The capacitance of a capacitor is given by:

$C = \frac{k\epsilon_o A}{d}$

k = dielectric constant

d = distance between the plates and A = area of the plates

so, the capacitance for this circular capacitor will be:

$C = \frac{\epsilon_o \pi r^2}{d}$

C = \frac{\epsilon_o \pi r^2}{d}

b]

From Ampere's law,

$\int B.dl = \mu_o I = \epsilon_o \mu_o \frac{d\phi_E}{dt}$

\int B.dl = \mu_o I = \epsilon_o \mu_o \frac{d\phi_E}{dt}

and the electric flux is the product of electric field and the area

so, $I = \epsilon_o \pi r^2\frac{dE}{dt}$

I = \epsilon_o \pi r^2\frac{dE}{dt}

so, the rate of change of electric field will be:

$\frac{dE}{dt}= \frac{I}{\epsilon_o \pi r^2}$

\frac{dE}{dt}= \frac{I}{\epsilon_o \pi r^2}