Question

(I) How much charge flows from each terminal of a 12.0-V battery when it is connected to a \({\bf{5}}\;{\bf{.00 - \mu F}}\) capacitor?

Short answer

The charge flows from each terminalis\(6.00 \times {10^{ - 5}}\;{\rm{C}}\).

Step-by-step solution

Understanding the charge on a capacitor

The capacitor is a charge storage device that consists of two parallel plates.

When a capacitor is connected to a battery, charge flows from the battery to each plate. There is some positive charge on one plate and the same amount of negative charge on the other plate.

The amount of charge flows from the battery is given as:

\(Q = CV\)

Here, Q is the charge, C is the capacitance and V is the voltage of the battery.

Given Data

The voltage is,\(V = 12.0\;{\rm{V}}\)

The capacitance is, \(C = 5.00\;\mu {\rm{F}}\)

Evaluation of the charge flows from each terminal

The relation to finding chargeis given by,

\(Q = CV\)

Substitute the values in the above expression.

\(\begin{aligned}Q &= \left( {5.00\;\mu {\rm{F}} \times \frac{{{{10}^{ - 6}}\;{\rm{F}}}}{{1\;\mu {\rm{F}}}}} \right)\left( {12.0\;{\rm{V}}} \right)\\Q &= 6.00 \times {10^{ - 5}}\;{\rm{C}}\end{aligned}\)

Thus, the charge flows from each terminal is \(6.00 \times {10^{ - 5}}\;{\rm{C}}\).