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PHYS1022 Electricity and Magnetism
Problem Sheet 6: tutorials
1. Figure 25.2 (W p 332) illustrates a mechanical analogue of a resistance in a simple
electric circuit – balls rolling down a hill interact with spikes before “a battery” lifts
them back to the top. Devise another mechanical analogue in which the current is
represented by a flow of water instead of balls rolling down a hill.
2. A constant electric field generally produces a constant drift velocity in a conduction
wire. How is this consistent with Newton’s third law, i.e. that force results in
acceleration, not velocity?
3. The figure shows an open switch, a battery of potential difference V, a currentmeasuring meter A, and three identical uncharged capacitors of capacitance C. When
the switch is closed and the circuit reaches equilibrium, what are
(a) the potential difference across each capacitor and
(b) the charge on the left-hand plate of each capacitor?
Give your answers in terms of C and V.
Discuss what would happen to answers (a) and (b) if the middle capacitor was now
2C.
4. One capacitor is charged until its stored energy is 4.0 J. A second identical but
uncharged capacitor is then connected to it in parallel.
(a) If the charge distributes equally, what is now the total energy stored in the
electric fields?
(b) Where did the excess energy go?
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Solutions
1. Consider viscous water flowing through a series of pipes. A pump represents the battery
and the diameter of the pipes their resistance.
2. The electrons are accelerated by the electric field but then collide with ions in the metal
loosing energy to the ions. The result is a net mean velocity for the electrons.
3. (a) The charge on each plate of a capacitor should be equal and opposite - +Q on the left
here and –Q on the right.
“C = Q/V”
so the voltage is the same for each capacitor and must equal V/3
(b) Q = + CV/3
If we now make the central capacitor 2C then the voltage across it will be ½ that on the
other two (the charges on the plates remain equal for each capacitor). The voltages
would therefore be
2 V/5 ; V/5 ; 2 V/5
And the charge on the plates is 2 CV/5
4. The energy of a charged capacitor is given by U = Q2/2 C
After the two are connected there will be Q/2 charge on each and hence the energy will
be
U’ = 2 x (Q/2)2/2C = U/2 = 2.0 J
The excess energy is dissipated as heat in the wires and as electromagnetic energy.
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