Question

A 5 μF capacitor is charged from a 2 V battery. When fully charged how many electrons are on the negative plate?

(a) 3.13 × 10^13

(b) 1.56 × 10^13

(c) 6.25 × 10^13

(d) 1.25 × 10^14

Answer 1

Let's first find the charge of the capacitor:

$$Q=C\Delta V=5\cdot10^{-6}\ F\cdot2\ V=1.0\cdot10^{-5}\ C.$$

Finally, we can find the number of electrons on the negative plate:

$$Q=Ne,$$ $$N=\dfrac{Q}{e}=\dfrac{1.0\cdot10^{-5}\ C}{1.6\cdot10^{-19}\ C}=6.25\cdot10^{-19}\ electrons.$$

Therefore, the correct answer is (c).

Answer:

(c) 6.25 × 10^13.