Question

A \[600pF\]capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged \[600pF\]capacitor. What are the common potential (in V) and energy lost (in J) after reconnection?
(A) 100, \[6 \times {10^{ - 6}}\]
(B) 200, \[6 \times {10^{ - 5}}\]
(C) 200, \[5 \times {10^{ - 6}}\]
(D) 100, \[6 \times {10^{ - 5}}\]

Answer 1

Hint: In this question, we need to determine the common potential (in V) and energy lost (in J) after reconnection of the uncharged capacitor with the battery. For this, we will use the relation between the capacitor, voltage supply and energy stored in a capacitor.

Complete step by step answer:
The capacitance of the capacitor \[C = 600pF = 600 \times {10^{ - 12}}F\]
Initial potential difference\[V = 200V\]
We know that the energy stored in a capacitor is given by the formula
\[E = \dfrac{1}{2}C{V^2} - - (i)\]
Now substitute the value of capacitance and the voltage in the equation (i); hence we get
\[
E = \dfrac{1}{2}C{V^2} \\
\Rightarrow E= \dfrac{1}{2} \times 600 \times {10^{ - 12}} \times {\left( {200} \right)^2} \\
\Rightarrow E= 1.2 \times {10^{ - 5}}J \\
 \]
So the energy stored in the capacitor when it was connected to the supply is\[ = 1.2 \times {10^{ - 5}}J\]
Now it is said that the capacitor is disconnected from the supply and it is connected to another uncharged \[600pF\]capacitor; hence the effective capacitance will be
\[
{C_{effective}} = \dfrac{{600 \times 600}}{{600 + 600}} \\
\Rightarrow{C_{effective}}= \dfrac{{360000}}{{1200}} \\
\Rightarrow{C_{effective}}= 300pF \\
 \]
So the energy stored in a capacitor will become
\[E' = \dfrac{1}{2}C'{V^2} - - (ii)\]
Now substitute the value of effective capacitance and the voltage in the equation (ii), we get
\[
E' = \dfrac{1}{2}C'{V^2} \\
\Rightarrow E’= \dfrac{1}{2} \times \left( {300 \times {{10}^{ - 12}}} \right) \times 200 \times 200 \\
\Rightarrow E’= 0.6 \times {10^{ - 5}}J \\
\]
Hence the loss in stored energy will be
\[
\Rightarrow E - E' \\
\Rightarrow 1.2 \times {10^{ - 5}} - 0.6 \times {10^{ - 5}} \\
\Rightarrow 6 \times {10^{ - 6}}J \\
 \]
Let the capacitor which was charged be\[{Q_1}\]and the uncharged capacitor be\[{Q_2}\]and\[{Q'_1}\], \[{Q'_2}\]be charge after they are connected together; hence we can write
\[{Q_1} + {Q_2} = {Q'_1} + {Q'_2} - - (iii)\]
Where \[{Q_2} = 0\]initially it was not connected, hence we can write equation (iii) as
\[
{Q_1} + {Q_2} = {{Q'}_1} + {{Q'}_2} \\
\Rightarrow{C_1} \times 200 + 0 = {C_1} \times {V_C} + {C_2} \times {V_C} \\
\Rightarrow{V_C}\left( {{C_1} + {C_2}} \right) = {C_1} \times 200 \\
\Rightarrow{V_C} = \dfrac{{{C_1}}}{{{C_1} + {C_2}}} \times 200 \\
 \]
Hence by substituting the value of capacitance we get
\[
{V_C} = \dfrac{{{C_1}}}{{{C_1} + {C_2}}} \times 200 \\
\Rightarrow{V_C} = \dfrac{{600}}{{600 + 600}} \times 200 \\
\Rightarrow{V_C} = 100V \\
 \]
Hence the common potential \[{V_C} = 100V\] and option A is the correct.

Note: It is interesting to note here that the equivalent capacitance follows the inverse rule of the resistances connected in series and/or parallel connections. Moreover, the product of the capacitance of the capacitor and the voltage across it results in charge of the capacitor.