. . . larger?¹

At point A. (Twice as strong as at B.)

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

. . . zero?²

Positive. You are moving ``against the field'' and hence ``uphill.''

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

. . . B?³

Zero. If you come in along the vertical axis, the field is always perpendicular to your motion and therefore does no work.

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

. . . given.⁴

If the plates carry a charge Q then the surface charge density is $\sigma = Q/A$. The field between the plates would thus be $E = \sigma/\epsilon_\circ = Q/\epsilon_\circ A$ if the gap were empty. The effect of the dielectric is to reduce E by a factor $\epsilon_\circ / \epsilon$. Thus in region 1 the field is $E_1 = Q/\epsilon_1 A$ and in region 2 the field is $E_2 = Q/\epsilon_2 A$. Moving from one plate to the other we travel a distance d/2 along E₁ and another distance d/2 along E2 for a net potential change of $V = E_1 d/2 + E_2 d/2 = (Qd/2A)(1/\epsilon_1 + 1/\epsilon_2)$. Since C is defined by Q = CV or C = Q/V, we have .

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

. . . $Q_\circ/e$?⁵

I.e. what is the time constant $\tau = R_{\rm eff} C_{\rm eff}$ of this circuit?

By symmetry the same charge must flow through each of the resistors onto each of the capacitors. The three resistors in the lower left corner are thus in parallel and have an effective resistance of 1/3 $\Omega$; similarly for the three resistors in the upper right corner. These two effective resistors are in series, giving an overall $R_{\rm eff}$ = 2/3 $\Omega$. By similar logic, the capacitors are all in parallel, giving immediately $C_{\rm eff} = 6 C = 6$ F. Thus $\tau = ({2\over3} \; \Omega) \times$6 F) or .

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

. . . diagram.⁶

3.0in

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

. . . refrigerator?⁷

An engine. More work ( $\int p \, dV$) is done by the gas in the expansion (3) at high temperature than is done on the gas in the compression (1) at low temperature. No work is done in steps 2 and 4.

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

. . . Earth.⁸

A charge Q will distribute itself uniformly over the surface of a conducting spherical shell (all the charges try to get as far away from each other as possible). The resultant charge distribution has spherical symmetry, giving an isotropic radial electric field (outside the sphere) that is just like that from a point source. The electrostatic potential (relative to any point at infinite distance) is therefore ${\displaystyle V = k_E {Q \over R_E} }$. The capacitance C = Q/V is therefore ${\displaystyle C_E = {R_E \over k_E} = {6.367 \times 10^6 \over 8.988 \times 10^9} }$ or . (Pretty small, when you consider that you can buy a 5 F capacitor that fits in the palm of your hand!)

(This is really a ``Quickie'' with numbers, meant as a gift.)

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

. . . ).⁹

GAUSS' LAW can be combined with cylindrical symmetry to show first that (there is no charge enclosed within a cylindrical surface whose radius is less that a). Next consider r>b: all the charge in a segment of length L is enclosed within a cylindrical surface of radius r>b, so the result is the same as for a line of charge: $\epsilon_\circ E(r>b) \times 2 \pi r L = L \left( \pi b^2 - \pi a^2 \right) \rho$ or . Finally, for a<r<b the enclosed charge in a segment of length L is $L \left( \pi r^2 - \pi a^2 \right) \rho$, giving .

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.