Question

A \[10.0\,\mu F\] parallel-plate capacitor with circular plates is connected to a
\[12.0\,V\] battery. How much charge (in \[\mu C\]) would be on the plates if their separation were doubled while the capacitor remained connected to the battery?

Answer 1

Hint: Determine the charge on the plates of the capacitor at initial separation between the plates using the relation between capacitance and potential difference. Recall the dependence of the capacitance of the capacitor with separation between the plates.

We can use the Formula:
\[Q = CV\]

Here, Q is the charge, C is the capacitance and V is the potential difference.

\[C = {\varepsilon _0}\dfrac{A}{d}\]

Here, \[{\varepsilon _0}\] is the permittivity of vacuum, A is the area of the plates of the capacitor and d is the distance between the plates.


Step by step answer: We have given that the potential difference across the plates of the capacitor is 12.0 V.

The charge on the capacitor plate is the product of capacitance of the capacitor and potential difference across the plates.

Therefore,
\[Q = CV\]

Here, Q is the charge, C is the capacitance and V is the potential difference.

Substitute \[10.0\,\mu F\] for C, and \[12.0\,V\] for V in the above equation.
\[Q = \left( {10.0\,\mu F} \right)\left( {12.0\,V} \right)\]
\[Q = 120\,\mu C\]

Now, on doubling the distance between the plates of the capacitor, the capacitance of the capacitor will be changed.

The capacitance of the capacitor is given as,
\[C = {\varepsilon _0}\dfrac{A}{d}\]

Here, \[{\varepsilon _0}\] is the permittivity of vacuum, A is the area of the plates of the capacitor and d is the distance between the plates.

From the above equation, we can conclude,
\[C \propto \dfrac{1}{d}\]

Therefore, we can write,
\[\dfrac{{{C_2}}}{{{C_1}}} = \dfrac{{{d_1}}}{{{d_2}}}\]
\[ \Rightarrow {C_2} = {C_1}\dfrac{{{d_1}}}{{{d_2}}}\]

Substitute \[10.0\,\mu F\] for \[{C_1}\] and \[2{d_1}\] for \[{d_2}\] in the above equation.
 \[{C_2} = \left( {10.0\,\mu F} \right)\dfrac{{{d_1}}}{{2{d_1}}}\]
\[\therefore {C_2} = 5\,\mu F\]

Now, the charge on the capacitor of capacitance \[5\,\mu F\] is,
 \[{Q_2} = \left( {5.0\,\mu F} \right)\left( {12.0\,V} \right)\]
\[{Q_2} = 60.0\,\mu C\]

Therefore, the charge on the capacitor plates after doubling the separation will be \[60.0\,\mu C\].


Note: Remember, after changing the separation between the plates of the capacitor, the surface area of the plates does not change. Therefore, we have taken the proportionality only with the separation of the plates. A farad is the unit of the capacitance which is very large, therefore, generally the capacitance is given in microfarads.