1. Two identical batteries, A and B, each have an emf of 1.5 V. The positive ter
ID: 1895267 • Letter: 1
Question
1. Two identical batteries, A and B, each have an emf of 1.5 V. The positive terminal of battery A is connected via a 10 cm long piece of copper wire, 0.50 mm in diameter, to one plate of a capacitor. The negative terminal is connected by an identical wire to the other plate of the capacitor. The capacitor has plates of surface area 20 cm2 separated by 0.10 mm. The positive terminal of battery B is connected directly to the negative terminal of battery B via a copper wire, with 0.50 mm diameter and a length of 15 m.(a) Battery A has been connected to the capacitor for long enough that the capacitor is fully charged. Find the electric field strengths at each of
i. a point inside the wire, halfway between the positive terminal of battery A and the capacitor
ii. a point inside the capacitor iii. a point inside the wire, halfway between the positive and negative terminals of
battery B
(b) Find the currents in each wire.
(c) A common belief is that batteries produce current. Based on the situation in this question, comment on this belief.
Explanation / Answer
(a)i. There is no current in the wire, so there is no electric field. 0.
ii. Inside the capacitor, the field is given by V/d = 1.5/10^-4 = 15,000 N/C.
iii. Halfway between the terminals of B, the field is V/d = 1.5/15 = .1 N/C.
(b) Both wires in circuit A carry no current, and Circuit B carries V/R = V/(l/a) = 1.5/(15*(1.68 x 10^-8)/(*.00025^2)) = 1.17 A.
(c) Circuit A has no current in the presence of a battery, so a more correct statement would be that a battery supplies a potential difference.
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