Question

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 12 micro C flows to the positive plate. The 4.0 V battery is then disconnected and replaced with a 5.0 V battery, with the positive and negative terminals connected in the same manner as before. How much additional charge flows to the positive plate?

Answer 1

Concepts and reason

The concept required to solve the given problem is of the principle of the capacitor and the charge flow. Initially, use the principle of the capacitor to get the relation of charge and applied voltage. Then take the ratio of two charges to a particular voltage to find the value of charge when a battery is connected. Then, to find the additional charge, take the difference of charge obtained when another battery is connected, and the charge obtained when a battery is connected.

Fundamentals

 

According to the principle of capacitors, \(C=\frac{Q}{V} \ldots \ldots(1)\)

Here, \(\mathrm{C}\) is the capacitance, \(\mathrm{Q}\) is the charge, and \(\mathrm{V}\) is the applied voltage.

 

According to the equation (1), if \(C\) is constant, then \(Q\) is directly proportional to V. Thus, \(Q \propto V\)

So,

$$ \frac{Q_{1}}{Q_{2}}=\frac{V_{1}}{V_{2}} $$

Here, \(Q_{\text {is the charge when } 4.0 ~ V ~ \text { battery is connected, } V_{1} \text { is the } 4.0 \mathrm{~V} \text { voltage, } Q_{2} \text { is the charge when } 5.0 \mathrm{~V} \text { battery is }}\) connected, and \(V_{2}\) is the \(5.0 \mathrm{~V}\) voltage.

Substitute \(12 \mu \mathrm{C}\) for \(Q_{1}, 4.0 \mathrm{~V}\) for \(V_{1},\) and \(5.0 \mathrm{~V}\) for \(V_{2}\) in equation \((2) .\)

\(\frac{12 \mu \mathrm{C}}{Q_{2}}=\frac{4.0 \mathrm{~V}}{5.0 \mathrm{~V}}\)

Rearrange the equation to find the value of \(Q_{2}\). \(Q_{2}=\frac{(12 \mu \mathrm{C})(5.0 \mathrm{~V})}{(4.0 \mathrm{~V})}\)

\(=15 \mu \mathrm{C}\)

Here, the capacitor connected is same. Thus, the value of charge will only depend on the voltages and the previous charge.

 

The charge obtained when \(4.0 \mathrm{~V}\) battery is connected is \(12 \mu \mathrm{C}\). The charge obtained when \(5.0 \mathrm{~V}\) battery is

connected is \(15 \mu \mathrm{C}\). Thus,

$$ Q=Q_{2}-Q_{1} $$

Substitute \(12 \mu \mathrm{C}\) for \(Q_{1}\) and \(15 \mu \mathrm{C}\) for \(Q_{2}\) in the above equation.

$$ Q=15 \mu \mathrm{C}-12 \mu \mathrm{C} $$

\(=3.0 \mu \mathrm{C}\)

The additional charge flows to the positive plate is of \(3.0 \mu \mathrm{C}\)

The additional charge is calculated when the initial charge available on the plate is get subtracted by the final charge. So, the value will be the additional charge added.