Question

Derive the formula for energy loss on connecting two charged conductors through a conducting wire.
OR
Explain the meaning of polarization of a dielectric material. What will be the effect on the capacity of an air filled parallel plate capacitor on putting a dielectric material in between its plates? Equivalent capacity of two condensers in series is 7.5 µF and in parallel is 30 µF. What will be the capacity of both the capacitors separately?

Hint:

Energy loss = (initial stored energy) - (energy after reaching common potential); it reduces to \(C_1C_2(V_1-V_2)^2/[2(C_1+C_2)]\). For the capacitors use \(C_1+C_2=30\) and \(C_1C_2=7.5\times30=225\).

Answer 1

Option 1: Energy loss on joining two charged conductors

Step 1 (Setup): Take two conductors of capacitances \(C_1\) and \(C_2\) charged to potentials \(V_1\) and \(V_2\) respectively (let \(V_1 > V_2\)). Their initial charges are \(q_1=C_1V_1\) and \(q_2=C_2V_2\).

Step 2 (Initial energy): Energy stored in a capacitor is \(U=\tfrac{1}{2}CV^2\). So total initial energy is
\[ U_i=\tfrac{1}{2}C_1V_1^{2}+\tfrac{1}{2}C_2V_2^{2}. \]
Step 3 (Common potential): When joined by a wire, charge flows until both reach a common potential \(V\). Charge is conserved:
\[ (C_1+C_2)V=C_1V_1+C_2V_2 \;\Rightarrow\; V=\frac{C_1V_1+C_2V_2}{C_1+C_2}. \]
Step 4 (Final energy): The two conductors now act as one capacitor \((C_1+C_2)\) at potential \(V\):
\[ U_f=\tfrac{1}{2}(C_1+C_2)V^{2}=\frac{(C_1V_1+C_2V_2)^{2}}{2(C_1+C_2)}. \]
Step 5 (Energy loss): \(\Delta U=U_i-U_f\). Putting both over the common denominator \(2(C_1+C_2)\):
\[ \Delta U=\frac{(C_1V_1^{2}+C_2V_2^{2})(C_1+C_2)-(C_1V_1+C_2V_2)^{2}}{2(C_1+C_2)}. \]
Expanding the numerator, the \(C_1^2V_1^2\) and \(C_2^2V_2^2\) terms cancel, leaving \(C_1C_2V_1^{2}+C_1C_2V_2^{2}-2C_1C_2V_1V_2=C_1C_2(V_1-V_2)^{2}\).
\[ \boxed{\;\Delta U=\frac{C_1C_2\,(V_1-V_2)^{2}}{2(C_1+C_2)}\;} \]
Since this is a positive quantity, energy is always lost (as heat in the connecting wire) unless \(V_1=V_2\).

Option 2: Polarization and the capacitor problem

Step 1 (Polarization): A dielectric has no free charges, but its molecules contain bound positive and negative charges. On placing the dielectric in an external field \(E_0\), the positive charges shift slightly along the field and negative charges opposite to it. Each molecule becomes a tiny dipole, and equal and opposite bound charges appear on the two faces of the slab. This alignment of molecular dipoles is called polarization.

Step 2 (Effect on capacitance): The bound surface charges set up an internal field \(E_p\) opposite to \(E_0\), so the net field inside becomes \(E=E_0-E_p=E_0/K\), where \(K\) is the dielectric constant \((K>1)\). For a fixed charge the potential difference \(V=Ed\) falls by factor \(K\), so the capacitance rises:
\[ C=\frac{K\,\varepsilon_0 A}{d}=K\,C_0. \]
Hence inserting a dielectric increases the capacity by the factor \(K\).

Step 3 (Series and parallel data): Let the two capacitors be \(C_1\) and \(C_2\).
Parallel: \(C_1+C_2=30\;\mu\text{F}\).
Series: \(\dfrac{C_1C_2}{C_1+C_2}=7.5\;\mu\text{F}\).
Step 4 (Solve): From the series relation, \(C_1C_2=7.5\times(C_1+C_2)=7.5\times30=225\).
So \(C_1\) and \(C_2\) are roots of \(x^{2}-(C_1+C_2)x+C_1C_2=0\):
\[ x^{2}-30x+225=0. \]
Discriminant \(=30^{2}-4(225)=900-900=0\), so \(x=\dfrac{30}{2}=15\) (repeated root).
Step 5 (Check): With \(C_1=C_2=15\;\mu\text{F}\): parallel \(=15+15=30\;\mu\text{F}\) & series \(=\dfrac{15\times15}{30}=7.5\;\mu\text{F}\). Both match.
\[ \boxed{\;C_1=C_2=15\;\mu\text{F}\;} \]