JEE Physics Electrostatic Potential and Capacitance MCQs Set 05

Answer: (a) 32

Explanation:
1. Each series string must share \(1000\ \text{V}\) with at most \(300\ \text{V}\) per capacitor, so at least \(\frac{1000}{300} = 3.33\), that is 4 capacitors per string (\(250\ \text{V}\) each).
2. A string of 4 has capacitance \(0.25\ \mu\text{F}\).
3. Strings in parallel add: \(\frac{2}{0.25} = 8\) strings.
4. Total \(= 4 \times 8 = 32\).

Teacher's Note:
1. Fix the number in series from the voltage rating and the number of parallel strings from the capacitance needed.
2. Round the series count up, never down.

Answer: (b) \(100\ \mu\text{C}\)

Explanation:
1. In the bridge, \(\frac{30}{60} = \frac{15}{30}\), so the bridge is balanced and the \(80\ \mu\text{F}\) capacitor carries no charge.
2. Top arm: \(30\) and \(60\) in series gives \(20\ \mu\text{F}\); bottom arm: \(15\) and \(30\) in series gives \(10\ \mu\text{F}\). Bridge \(= 30\ \mu\text{F}\).
3. Bridge in series with the other \(30\ \mu\text{F}\): \(C_{eq} = 15\ \mu\text{F}\), total charge \(= 15\times 10 = 150\ \mu\text{C}\).
4. Potential difference across the bridge \(= \frac{150}{30} = 5\ \text{V}\).
5. Charge on the top arm (and on the \(60\ \mu\text{F}\) capacitor) \(= 20 \times 5 = 100\ \mu\text{C}\).

Teacher's Note:
1. Check the bridge condition first; a balanced bridge lets you ignore the middle capacitor.
2. Capacitors in series carry the same charge, so find the arm charge from the arm voltage.

Answer: (a) \(0.92\ \text{ms}\)

Explanation:
1. \(q = q_0(1 - e^{-t/RC})\) with \(RC = 2 \times 100\times 10^{-6} = 2\times 10^{-4}\ \text{s}\).
2. For \(q = 0.99\,q_0\): \(e^{-t/RC} = 0.01\), so \(t = RC\ln 100 = 2\times10^{-4}\times 4.605\).
3. \(t \approx 0.92\times 10^{-3}\ \text{s} = 0.92\ \text{ms}\).

Teacher's Note:
1. The charging law is exponential, so the time to reach a given fraction uses the natural logarithm.
2. \(\ln 100 \approx 4.6\) is worth remembering.

Answer: (b) \(0.9\ \text{A}\)

Explanation:
1. In the steady state no current flows in the branch with the capacitor.
2. The \(2\ \Omega\) and \(3\ \Omega\) resistors in parallel give \(\frac{2\times 3}{5} = 1.2\ \Omega\).
3. Current from the battery \(I = \frac{6}{2.8 + 1.2} = 1.5\ \text{A}\).
4. Voltage across the parallel pair \(= 1.5 \times 1.2 = 1.8\ \text{V}\).
5. Current in the \(2\ \Omega\) resistor \(= \frac{1.8}{2} = 0.9\ \text{A}\).

Teacher's Note:
1. A fully charged capacitor blocks steady current, so delete its branch from the circuit.
2. Then use parallel and series rules and finally the current-division idea.

Answer: (a) \(\frac{3V}{K+2}\)

Explanation:
1. Total charge \(= (2C + C)V = 3CV\) stays constant after the battery is removed.
2. New capacitances: \(2C\) and \(KC\), still in parallel, total \((K+2)C\).
3. \(V' = \frac{3CV}{(K+2)C} = \frac{3V}{K+2}\).

Teacher's Note:
1. After disconnecting the battery, total charge is conserved when the capacitors stay joined.
2. Parallel capacitors share one common potential.

Answer: (d) \(3\ \text{V}\)

Explanation:
1. Series capacitors carry the same charge: \(Q = 6\times 2 = 12\ \mu\text{C}\).
2. Voltage across the \(12\ \mu\text{F}\) capacitor \(= \frac{12}{12} = 1\ \text{V}\).
3. Battery voltage \(= 2 + 1 = 3\ \text{V}\).

Teacher's Note:
1. In a series combination the charge is common and the voltage divides inversely with capacitance.
2. The smaller capacitor takes the larger share of the voltage.

Answer: (a) \(\frac{7}{8}C\)

Explanation:
1. Let the lower nodes be \(a\), \(m\), \(b\) (with \(x\) joined to \(a\) and \(y\) to \(b\)) and the upper nodes be \(P\) (above \(a\)) and \(Q\) (above \(b\)).
2. Apply \(x = a\) at potential 1 and \(y = b\) at 0; by symmetry \(m\) is at potential \(\frac{1}{2}\) and \(P\), \(Q\) have potentials \(p\) and \(1-p\).
3. Charge balance at \(P\): \((1-p) + (\tfrac{1}{2}-p) + (1-2p) = 0\), so \(p = \frac{5}{8}\).
4. Charge leaving \(x\) \(= C\left[(1-\tfrac{5}{8}) + (1-\tfrac{1}{2})\right] = \frac{7}{8}C\) for unit potential difference.
5. Hence \(C_{eq} = \frac{7}{8}C\).

Teacher's Note:
1. When a network has mirror symmetry, symmetric nodes sit at symmetric potentials.
2. Write charge balance at the unknown node and solve one linear equation.

Answer: (d) \(0.32\ \text{J}\)

Explanation:
1. All the stored energy is dissipated as heat in the resistor.
2. \(U = \frac{1}{2}CV^2 = \frac{1}{2}(4\times10^{-6})(400)^2 = 0.32\ \text{J}\).

Teacher's Note:
1. The value of the resistance does not change the total heat; it only changes how fast the heat is produced.
2. Heat equals the initial stored energy.

Answer: (c) \(\frac{3\varepsilon_0 A}{d}\)

Explanation:
1. The three gaps (1-2, 2-3 and 3-4) each form a capacitor of capacitance \(\frac{\varepsilon_0 A}{d}\).
2. Plates 1 and 3 are at one potential and plates 2 and 4 at the other, so every gap has the same potential difference and the three capacitors are in parallel.
3. \(C = \frac{3\varepsilon_0 A}{d}\).

Teacher's Note:
1. Count the gaps between facing plates, not the plates.
2. Alternate plates joined together make the gap capacitors parallel.

Answer: (d) \(8\ \text{mm}\)

Explanation:
1. Original \(C_0 = \frac{\varepsilon_0 A}{4}\).
2. With the slab: \(C = \frac{\varepsilon_0 A}{d' - t + t/K} = \frac{\varepsilon_0 A}{d' - 3 + 1} = \frac{\varepsilon_0 A}{d' - 2}\).
3. Set \(C = \frac{2}{3}C_0\): \(d' - 2 = \frac{3\times4}{2} = 6\), so \(d' = 8\ \text{mm}\).

Teacher's Note:
1. A slab of thickness t and constant K acts like an air gap of \(\frac{t}{K}\).
2. So the effective separation becomes \(d - t + \frac{t}{K}\).

Answer: (d) a decrease in the energy of the system unless \(Q_1R_2 = Q_2R_1\)

Explanation:
1. Charge flows until both spheres reach the same potential, and some energy is lost as heat during the flow.
2. If the spheres are already at the same potential, \(\frac{Q_1}{R_1} = \frac{Q_2}{R_2}\), that is \(Q_1R_2 = Q_2R_1\), no charge flows and the energy does not change.
3. Otherwise the energy decreases.

Teacher's Note:
1. Energy is lost only when charge actually flows between the two conductors.
2. Equal potentials mean no flow and no loss.

Answer: (d) the metal of the plates

Explanation:
1. \(C = \frac{K\varepsilon_0 A}{d}\) depends on the area, the separation and the medium between the plates.
2. It does not depend on the metal the plates are made of.

Teacher's Note:
1. Capacitance is purely geometric together with the medium between the conductors.
2. The charge and potential do not change it either.

Answer: (b) \(\frac{1}{3}\ \text{nC}\)

Explanation:
1. The field at the surface is \(E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\), and it must not exceed \(3\times10^6\ \text{V/m}\).
2. \(q = \frac{E r^2}{9\times10^9} = \frac{3\times10^6\times10^{-6}}{9\times10^{9}} = 3.3\times10^{-10}\ \text{C}\).
3. This is \(\frac{1}{3}\ \text{nC}\).

Teacher's Note:
1. Air breaks down when the field at the surface exceeds its dielectric strength.
2. Use the surface field of a charged sphere to get the maximum charge.

Answer: (b) \(0.30\ \text{mJ}\)

Explanation:
1. Before closing S: two capacitors in parallel give \(4\ \mu\text{F}\), in series with \(2\ \mu\text{F}\): \(C_1 = \frac{4}{3}\ \mu\text{F}\), charge \(Q_1 = \frac{4}{3}\times 30 = 40\ \mu\text{C}\), energy \(= \frac{1}{2}\times\frac{4}{3}\times 900 = 600\ \mu\text{J}\).
2. After closing S the parallel pair is short-circuited, leaving \(2\ \mu\text{F}\) across the battery: \(Q_2 = 60\ \mu\text{C}\), energy \(= \frac{1}{2}\times 2\times 900 = 900\ \mu\text{J}\).
3. Work done by the battery \(= V\,\Delta Q = 30\times 20 = 600\ \mu\text{J}\).
4. Heat \(= W_{\text{battery}} - \Delta U = 600 - 300 = 300\ \mu\text{J} = 0.30\ \text{mJ}\).

Teacher's Note:
1. Heat in a capacitor circuit equals battery work minus the gain in stored energy.
2. Find the equivalent capacitance before and after the change.

Answer: (a) \(2C_0\)

Explanation:
1. Let A be at \(V\) and B at \(0\). Let the potentials of the nodes after the first \(2C_0\) and after the middle \(C_0\) be \(x\) and \(y\).
2. Charge balance at the first node: \(2C_0(V - x) = C_0(x - y) + 4C_0 x\).
3. Charge balance at the second node: \(C_0(x-y) + C_0(V - y) = 2C_0 y\).
4. Solving gives \(x = y = \frac{V}{3}\) (the middle \(C_0\) carries no charge).
5. Charge from A \(= 2C_0\cdot\frac{2V}{3} + C_0\cdot\frac{2V}{3} = 2C_0V\), so \(C_{eq} = 2C_0\).

Teacher's Note:
1. Assign potentials to the internal nodes and write charge balance for each.
2. If two nodes turn out to be at equal potential, the capacitor between them can be ignored.

Answer: (a) \(w + w_b = \Delta U\)

Explanation:
1. The battery has zero resistance and the process is very slow, so there is no heat loss: \(H = 0\).
2. By energy conservation the work of the external agent plus the work of the battery equals the change in the stored energy: \(w + w_b = \Delta U\).
3. The plates attract each other, so the agent does positive work \(w \neq 0\).

Teacher's Note:
1. Apply the work-energy principle to the whole system including the battery.
2. A slow process with a zero-resistance circuit produces no heat.

Answer: (d) \(30\ \text{V}\)

Explanation:
1. The capacitor \(C\) joined directly between A and B has no effect on the other branch.
2. The branch \(2C\), \(C\), \(2C\) is in series: \(\frac{1}{C_s} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{2C} = \frac{2}{C}\), so \(C_s = \frac{C}{2}\).
3. Charge \(= \frac{C}{2}\times 60 = 30C\).
4. Potential difference across the middle \(C\) (between M and N) \(= \frac{30C}{C} = 30\ \text{V}\).

Teacher's Note:
1. In a series branch, every capacitor carries the same charge.
2. A capacitor connected directly across the supply does not change the voltage across the other branch.

Answer: (d) \(\frac{1}{4}C(V_1 - V_2)^2\)

Explanation:
1. Initial energy \(= \frac{1}{2}C(V_1^2 + V_2^2)\).
2. Common potential \(V = \frac{V_1 + V_2}{2}\), final energy \(= \frac{1}{2}(2C)\left(\frac{V_1+V_2}{2}\right)^2 = \frac{C}{4}(V_1+V_2)^2\).
3. Loss \(= \frac{C}{4}\left[2V_1^2 + 2V_2^2 - (V_1+V_2)^2\right] = \frac{C}{4}(V_1 - V_2)^2\).

Teacher's Note:
1. Joining like plates conserves charge and gives a common potential equal to the average here.
2. The loss is always proportional to the square of the potential difference.

Answer: (c) The \(1\ \mu\text{F}\) capacitor loses 50% charge sooner than the \(2\ \mu\text{F}\) capacitor

Explanation:
1. Both are charged to the battery emf \(V\), so the initial current is \(\frac{V}{R}\) in both circuits; statements a and b are false.
2. Discharge time constant \(\tau = RC\) is smaller for the \(1\ \mu\text{F}\) capacitor.
3. Charge halves at \(t = \tau\ln 2\), so the \(1\ \mu\text{F}\) capacitor loses 50% of its charge sooner.

Teacher's Note:
1. Initial current depends on the voltage and the resistance, not on the capacitance.
2. Time constant \(RC\) decides how quickly the charge decays.

Answer: (b) \(\left(\frac{\cos\beta}{\cos\alpha}\right)^2\)

Explanation:
1. The field is perpendicular to the plates, so the velocity component parallel to the plates stays constant: \(v_1\cos\alpha = v_2\cos\beta\).
2. \(\frac{v_1}{v_2} = \frac{\cos\beta}{\cos\alpha}\).
3. Kinetic energy is proportional to \(v^2\), so \(\frac{K_1}{K_2} = \left(\frac{\cos\beta}{\cos\alpha}\right)^2\).

Teacher's Note:
1. A force perpendicular to the plates cannot change the parallel component of velocity.
2. Use the conserved component to connect the two speeds.

Answer: (a) \((K+3)\frac{C}{4}\)

Explanation:
1. The capacitor splits into two parallel capacitors.
2. The part filled by the slab has area \(\frac{A}{4}\): capacitance \(\frac{K C}{4}\).
3. The air part has area \(\frac{3A}{4}\): capacitance \(\frac{3C}{4}\).
4. Total \(= \frac{KC}{4} + \frac{3C}{4} = (K+3)\frac{C}{4}\).

Teacher's Note:
1. When the dielectric fills part of the area, treat the two regions as parallel capacitors.
2. Each region has the same plate separation.

Answer: (d) \(1.25\ \text{J}\)

Explanation:
1. Initial energy \(= \frac{1}{2}(5\times10^{-6})\left[(2000)^2 + (1000)^2\right] = 12.5\ \text{J}\).
2. Common potential \(= \frac{2000+1000}{2} = 1500\ \text{V}\); final energy \(= \frac{1}{2}(10\times10^{-6})(1500)^2 = 11.25\ \text{J}\).
3. Loss \(= 12.5 - 11.25 = 1.25\ \text{J}\).
4. Check with the formula \(\frac{C}{4}(V_1-V_2)^2 = \frac{5\times10^{-6}}{4}(10^3)^2 = 1.25\ \text{J}\).

Teacher's Note:
1. Joining like terminals gives a common potential equal to the average for identical capacitors.
2. The energy lost is always a positive quantity.

Answer: (c) \(1 : 5\)

Explanation:
1. The \(1\ \mu\text{F}\) and \(3\ \mu\text{F}\) capacitors are in parallel between the same two nodes: \(4\ \mu\text{F}\).
2. Chain: \(2\), \(4\) and \(2\) in series: \(\frac{1}{C} = \frac{1}{2} + \frac{1}{4} + \frac{1}{2} = \frac{5}{4}\), \(C = 0.8\ \mu\text{F}\), charge \(= 0.8\times10 = 8\ \mu\text{C}\).
3. Potential difference across the \(4\ \mu\text{F}\) block (hence across the \(1\ \mu\text{F}\)) \(= \frac{8}{4} = 2\ \text{V}\).
4. The \(5\ \mu\text{F}\) capacitor is directly across the battery: \(10\ \text{V}\).
5. Ratio \(= 2 : 10 = 1 : 5\).

Teacher's Note:
1. Capacitors joined across the same two nodes are in parallel and share the same voltage.
2. The \(5\ \mu\text{F}\) capacitor is in parallel with the whole chain, so it sees the full emf.

Answer: (d) \(27\ \mu\text{C}\)

Explanation:
1. Switch open: the capacitors are in series, \(2\ \mu\text{F}\), charge \(= 2\times 9 = 18\ \mu\text{C}\) on each; the net charge on node X is zero.
2. The resistors carry a steady current \(\frac{9}{9} = 1\ \text{A}\), so the voltages across them are \(3\ \text{V}\) and \(6\ \text{V}\).
3. Switch closed: X is at the potential of Y, so the \(3\ \mu\text{F}\) capacitor has \(3\ \text{V}\) and the \(6\ \mu\text{F}\) has \(6\ \text{V}\). Charges: \(3\times3 = 9\ \mu\text{C}\) and \(6\times6 = 36\ \mu\text{C}\).
4. Net charge now on node X has magnitude \(36 - 9 = 27\ \mu\text{C}\), and it came through the switch.

Teacher's Note:
1. Compare the net charge on the isolated node X before and after the switch is closed.
2. In the steady state, current flows only through the resistors.

Answer: (b) 5

Explanation:
1. The same potential with the same charge means the capacitance is unchanged.
2. \(d' - t + \frac{t}{K} = d\) with \(d' = d + 1.6\) and \(t = 2\): \(1.6 - 2 + \frac{2}{K} = 0\).
3. \(\frac{2}{K} = 0.4\), so \(K = 5\).

Teacher's Note:
1. Equal capacitance requires equal effective air separation.
2. A slab of thickness t and constant K behaves as \(\frac{t}{K}\) of air.