28.08.2016 Views

Bukhovtsev-et-al-Problems-in-Elementary-Physics

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

310 ANSWERS AND SOLUTIONS<br />

+/l<br />

U<br />

-1/<br />

~u<br />

+1/-1/ -1/+1/<br />

8 r:<br />

C<br />

(b) (c) Fig. 425<br />

When the capacitor C 2 is short-circuited, the charge -q is neutr<strong>al</strong>ized.<br />

The potenti<strong>al</strong> difference b<strong>et</strong>ween the plates of the capacitor C should rema<strong>in</strong><br />

approximately equ<strong>al</strong> to U, s<strong>in</strong>ce q ~ Q. The work required to move the charge<br />

<strong>al</strong>ong the circuit ABCDA is zero. Therefore, the voltage across the capacitor<br />

C t should become equ<strong>al</strong> to ~ U and the charges on it to +2q and<br />

-2q. The charges on the plates of the capacitor C will be +Q -q and<br />

- Q + q, respectively (Fig. 425b).<br />

When the capacitor C is disconnected from the earth, the distribution of<br />

the charges and, hence, of the potenti<strong>al</strong>s will not change.<br />

If the capacitor C 1 is shorted, the charges will be redistributed as shown<br />

<strong>in</strong> Fig. 425c. It is only <strong>in</strong> this case that the required potenti<strong>al</strong>s relative to<br />

the earth will be obta<strong>in</strong>ed: zero on the left plate and ~ U on the right<br />

one. When the plates are earthed <strong>al</strong>ternately, the potenti<strong>al</strong> difference b<strong>et</strong>ween<br />

the plates will gradua lly drop because the charge decreases.<br />

440. No, they will not. When the plates are <strong>al</strong>ternately earthed the same<br />

processes will take place as <strong>in</strong> the absence of the battery (see Problem 439).<br />

The only difference is that the potenti<strong>al</strong> difference b<strong>et</strong>ween the plates is<br />

<strong>al</strong>ways kept constant.<br />

441. The tot<strong>al</strong> energy of the two capacitors before connection is<br />

1 (~ 2)<br />

Weo= 2 CIU1+C2U~<br />

and after connection<br />

W _..!..~_..!..(CIUl+C2U2)2<br />

er: 2 C 1+C2<br />

- 2 C1+C 9<br />

It is easy to see that Weo> We. The difference <strong>in</strong> the energies is<br />

Weo-We=c~f~s (U~+U:-2U1U2) > 0<br />

When U 1=U2 , we have W eo - We=O, and when C 1=C2 and U 2=O, then<br />

Weo=2We·<br />

The electrostatic energy dim<strong>in</strong>ishes because when the capacitors are connected<br />

by conductors, the charges flow from one capacitor to the other. Heat<br />

is liberated <strong>in</strong> the connect<strong>in</strong>g conductors. The quantity of heat evolved will<br />

not depend on the resistance of the conductors. When the resistance is low,<br />

the conductors will <strong>al</strong>low greater currents to flow through them, and vice<br />

versa.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!