Electrostatic Potential And Capacitance - 06 : Questions || Questions || Earthing

Main Ideas

1. Dielectric Strength Limits How Much Charge Can Be Stored

A conductor’s potential rises as more charge is added. If its electric field becomes too strong, the surrounding insulating medium can break down and begin conducting, allowing charge to escape.

The maximum charge is reached when the electric field at the conductor’s surface reaches the dielectric strength of the surrounding medium. For pure air, the lesson gives a value of approximately (3\times10^6\ \text{V/m}). Moisture and impurities can lower this value, which is why high-voltage conductors need sufficient separation.

Capacitance describes how much charge a conductor or capacitor can store for a given potential difference. No real setup can store unlimited charge.

Examples

• Sphere of radius (3\ \text{m}): Set its surface field, (E=kQ/R^2), equal to (3\times10^6\ \text{V/m}). The maximum charge is (Q=3\times10^{-3}\ \text{C}), or 3 mC.
• Parallel plates (5\ \text{mm}) apart: Using (E=V/d), the maximum voltage in air is [ V=(3\times10^6)(5\times10^{-3})=\mathbf{15\ \text{kV}}. ] Reducing the gap reduces the maximum voltage before air breakdown. A dielectric material can be used instead of air to change the capacitor’s properties.

2. Capacitance of Different Geometries

The general approach is to identify the potential difference between the relevant conductors and use (C=Q/\Delta V).

Isolated Spherical Conductor

The field extends from the sphere’s surface to infinity. Its capacitance is

[ C=4\pi\varepsilon_0R. ]

A capacitance of (1\ \text{F}) would require an isolated sphere with a radius of about (9\times10^9\ \text{m}), illustrating that a farad is a very large unit. Practical capacitances are commonly expressed in smaller units such as microfarads or picofarads.

Concentric Spherical Capacitor

For an inner sphere of radius (a) and a grounded outer sphere of radius (b), the field lies between the spheres, and the outer conductor’s potential is zero. The capacitance is

[ C=4\pi\varepsilon_0\frac{ab}{b-a}. ]

In the limit (b\to\infty), this reduces to the isolated-sphere result.

Concentric Cylindrical Capacitor

For inner and outer radii (a) and (b), the field is confined to the region between the cylinders when the outer cylinder is grounded. For cylinder length (L),

[ C=\frac{2\pi\varepsilon_0L}{\ln(b/a)}. ]

The capacitance per unit length is

[ \frac{C}{L}=\frac{2\pi\varepsilon_0}{\ln(b/a)}. ]

If the outer cylinder is not grounded, the field outside may also need to be included in the potential calculation.

Parallel-Plate Capacitor

For plates of area (A), separation (d), and negligible edge effects, the capacitance in vacuum or air is

[ C=\frac{\varepsilon_0A}{d}. ]

Filling the space with a dielectric of relative permittivity (K) gives

[ C=\frac{K\varepsilon_0A}{d}=KC_0. ]

The lesson distinguishes the net charge of the two-plate capacitor, which is zero, from “charge on the capacitor,” which conventionally means the magnitude of the charge on either plate.

3. Energy Stored in a Capacitor

The equivalent expressions for stored energy are

[ U=\frac{Q^2}{2C}=\frac12CV^2=\frac12QV. ]

Two derivations are described:

• Work done while charging: Transferring a small charge (dq) when the capacitor already holds charge (q) requires (dW=V\,dq=(q/C)dq). Integrating from zero to (Q) gives (U=Q^2/(2C)).
• Energy density in the electric field: Use the energy density (u=\tfrac12\varepsilon_0E^2), then multiply by the field-filled volume for a uniform field.

A fully charged capacitor in a DC circuit is at steady state: it draws no current, so ideal resistors in that path have no voltage drop. During charging, however, current flows and resistance can dissipate energy as heat.

4. Capacitors in Series and Parallel

Series

Every capacitor carries the same magnitude of charge, and the total voltage is the sum of the individual voltages. Since (V_i=Q/C_i),

[ \frac1{C_{\rm eq}}=\frac1{C_1}+\frac1{C_2}+\cdots. ]

For two capacitors,

[ C_{\rm eq}=\frac{C_1C_2}{C_1+C_2}. ]

For (N) identical capacitors of capacitance (C),

[ C_{\rm eq}=\frac{C}{N}. ]

Voltage divides inversely with capacitance. For two capacitors,

[ V_1=\frac{C_2}{C_1+C_2}V,\qquad V_2=\frac{C_1}{C_1+C_2}V. ]

Parallel

Each capacitor has the same voltage, and the total charge is the sum of the individual charges. Thus,

[ C_{\rm eq}=C_1+C_2+\cdots. ]

For (N) identical capacitors,

[ C_{\rm eq}=NC. ]

Charge divides in proportion to capacitance. For two capacitors,

[ Q_1=\frac{C_1}{C_1+C_2}Q,\qquad Q_2=\frac{C_2}{C_1+C_2}Q. ]

Worked examples

• Three series capacitors of (3), (6), and (18\ \mu\text{F}) across (100\ \text{V}) divide the voltage in the ratio (6:3:1), giving (60), (30), and (10\ \text{V}).
• Four parallel capacitors of (2), (4), (8), and (16\ \mu\text{F}) share a total charge of (100\ \mu\text{C}) in proportion to (1:2:4:8).

5. Battery Work and Heat During Charging

For initially uncharged capacitors, the battery’s work is (W_{\rm battery}=QV), while the final energy stored is (U=\tfrac12QV). The difference is dissipated as heat in the charging circuit. Thus, in the idealized charging example discussed, half the battery’s supplied energy is stored and half becomes heat.

For three series capacitors of (2), (3), and (6\ \mu\text{F}) connected to (30\ \text{V}):

• (C_{\rm eq}=1\ \mu\text{F}), so the charge is (30\ \mu\text{C}).
• The energy in the (3\ \mu\text{F}) capacitor is (150\ \mu\text{J}).
• The total stored energy is (450\ \mu\text{J}).
• Battery work is (QV=900\ \mu\text{J}), so (450\ \mu\text{J}) is dissipated as heat.

6. Common Potential When Charged Capacitors Are Connected

When two charged capacitors are connected with their positive plates together and their negative plates together, charge flows until both reach the same final potential.

Conserving total charge gives

[ V_{\rm common}=\frac{C_1V_1+C_2V_2}{C_1+C_2}. ]

The energy lost during redistribution is dissipated as heat:

[ H=\frac12\frac{C_1C_2}{C_1+C_2}(V_1-V_2)^2. ]

The lesson emphasizes deriving these results from charge conservation and equal final voltages rather than relying only on memorized formulas.

Speakers and Sources Featured

• Main speaker: A male physics instructor teaching electrostatic potential, capacitance, and capacitor circuits.
• Students and class participants: The instructor addresses students by names including Aditya, Pushpendra, Prakhar, Rehan, Zakir, Nishant, and Samrit. The subtitles do not reliably identify which students speak in each moment.
• Referenced materials and sources: NCERT textbooks and DD One (Doordarshan’s national television channel) are mentioned in a brief classroom aside; neither is presented as a separate speaker.
• Channel/source: PW Solutions.