Option 1: Heat dissipated when two charged bodies share charge
Step 1: Consider spheres with capacitances \(C_1,C_2\) raised to potentials \(V_1,V_2\). Their stored electrostatic energies are \(U_1=\tfrac12 C_1V_1^2\) and \(U_2=\tfrac12 C_2V_2^2\), giving initial total \(U_i=U_1+U_2\).
Step 2: A wire lets charge migrate from higher to lower potential. Total charge \(Q=C_1V_1+C_2V_2\) stays fixed, and the flow stops only when both bodies sit at one potential \(V_c\). Because the combined plate system stores charge \(Q\) on capacitance \(C_1+C_2\), \(V_c=\dfrac{Q}{C_1+C_2}=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}\).
Step 3: The energy afterwards can be written directly in terms of charge, \(U_f=\dfrac{Q^2}{2(C_1+C_2)}=\dfrac{(C_1V_1+C_2V_2)^2}{2(C_1+C_2)}\).
Step 4: A neat shortcut: the migrated charge is \(\Delta q=\dfrac{C_1C_2}{C_1+C_2}(V_1-V_2)\), and the average driving potential difference across the wire is \(\tfrac12(V_1-V_2)\). The heat produced equals charge moved times mean potential difference, \(H=\Delta q\times\tfrac12(V_1-V_2)\).
Step 5: Substituting \(\Delta q\),
\[ H=\frac{C_1C_2}{2(C_1+C_2)}(V_1-V_2)^2 \]
which is exactly \(U_i-U_f\). The result is always positive, confirming energy is dissipated as heat and radiation in the wire.
\[ \boxed{\;\Delta U=\dfrac{C_1C_2(V_1-V_2)^2}{2(C_1+C_2)}\;} \]
Option 2: Dielectric polarization and finding the two capacitors
Step 1: Polarization means the ordering of a dielectric's molecular dipoles under an applied field. Non-polar molecules develop induced dipoles, while polar molecules (already dipoles) turn to line up with the field. The result is a bound negative charge on the face where the field enters and a bound positive charge where it leaves; per unit volume this is described by the polarization vector \(P\).
Step 2: These bound charges oppose the applied field, weakening the net field between the plates to \(E_0/K\). With the source keeping charge fixed, the reduced field lowers the voltage, and since \(C=Q/V\), the capacitance grows to \(K\) times its air value: \(C=KC_0\). Thus a dielectric raises the capacitor's capacity.
Step 3: Let the capacitances be \(a\) and \(b\). Given data: parallel combination \(a+b=30\) and series combination \(\dfrac{ab}{a+b}=7.5\).
Step 4: Substitute \(a+b=30\) into the series equation: \(\dfrac{ab}{30}=7.5\Rightarrow ab=225\). Now use \((a-b)^2=(a+b)^2-4ab=900-900=0\), so \(a-b=0\), i.e. \(a=b\).
Step 5: With \(a=b\) and \(a+b=30\), each is \(15\,\mu\)F. Verification of the series value: \(\dfrac{15\times15}{30}=7.5\,\mu\)F. Consistent.
\[ \boxed{\;C_1=C_2=15\,\mu\text{F}\;} \]