Question

(i) A parallel plate capacitor A of capacitance C is charged by a battery to a potential 'V'. The battery is disconnected and an uncharged identical capacitor B is connected across it. Calculate for the capacitor A the new value of the: (I) charge (II) potential difference (III) energy stored. Justify your answers.

(ii) Draw the pattern of electric field lines due to: (I) a positively charged conducting sphere, and (II) an electric dipole.

Hint:

Charge stays constant after the battery is removed. Identical capacitors share it equally, so each has half the charge and half the potential. Energy on A becomes one fourth.

Answer 1

Step 1: Understanding the situation:
Capacitor A has capacitance \(C\) and potential \(V\), so its charge is \(Q = CV\) and its energy is \(U = \tfrac{1}{2}CV^{2}\).
The battery is removed, so no charge can enter or leave the pair of capacitors. Then the identical capacitor B is joined across A.

Step 2: (i)(I) New charge on A:
The total charge \(CV\) is conserved. The two capacitors are identical, so charge flows until both have the same potential. Each then holds half of the total.
\[ Q_{A,n} = \frac{CV}{2} \]
Justification: identical capacitors at equal potential carry equal charge, and the sum is fixed.

Step 3: (i)(II) New potential difference:
\[ V_{A,n} = \frac{Q_{A,n}}{C} = \frac{CV/2}{C} = \frac{V}{2} \]
Justification: the charge has halved while the capacitance stays the same.

Step 4: (i)(III) New energy stored in A:
\[ U_{A,n} = \frac{1}{2}C\left(\frac{V}{2}\right)^{2} = \frac{CV^{2}}{8} \]
This is one fourth of the original \(\tfrac{1}{2}CV^{2}\). The two capacitors together hold \(CV^{2}/4\), which is half the original energy. The rest is lost as heat and radiation in the connecting wires.

Step 5: (ii)(I) Field lines of a positively charged conducting sphere:
The lines are straight radial lines. They start on the surface and point outward, leaving the surface at right angles to it.
There are no lines inside the conductor, because the field inside is zero. Outside, the pattern is the same as for a point positive charge at the centre. The lines spread out and get less dense with distance.

Step 6: (ii)(II) Field lines of an electric dipole:
The lines are curves that start on the positive charge and end on the negative charge.
The pattern is symmetric about the dipole axis. The line along the axis is straight, and no two lines cross each other.

Final Answer:
New charge on A is \(CV/2\), new potential is \(V/2\) and new energy is \(CV^{2}/8\). The field line patterns are as described above. \[ \boxed{Q_{n} = \frac{CV}{2},\quad V_{n} = \frac{V}{2},\quad U_{n} = \frac{CV^{2}}{8}} \]