Electrostatic Potential and Capacitance

Given:
- $q_1 = 5 \times 10^{-8}$ C (positive charge)
- $q_2 = -3 \times 10^{-8}$ C (negative charge)
- Distance between charges: $d = 16$ cm $= 0.16$ m

Concept: Electric potential due to a point charge is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$. By superposition, the net potential is the algebraic sum of potentials due to individual charges.

We look for points where $V = 0$, i.e., $V_1 + V_2 = 0$.

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Case 1: Point P between the two charges

Let P be at distance $x$ from $q_1$ (so at distance $16 - x$ from $q_2$).

$$\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{x} + \frac{q_2}{16-x}\right) = 0$$

$$\frac{5 \times 10^{-8}}{x} + \frac{-3 \times 10^{-8}}{16 - x} = 0$$

$$\frac{5}{x} = \frac{3}{16 - x}$$

$$5(16 - x) = 3x$$

$$80 = 8x \implies x = 10 \text{ cm}$$

So one point is 10 cm from $q_1$ (between the charges).

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Case 2: Point Q outside the two charges (on the side of $q_2$, the smaller charge)

Let Q be at distance $y$ from $q_1$ (so at distance $y - 16$ from $q_2$), with y > 16 cm.

$$\frac{5 \times 10^{-8}}{y} + \frac{-3 \times 10^{-8}}{y - 16} = 0$$

$$\frac{5}{y} = \frac{3}{y - 16}$$

$$5(y - 16) = 3y$$

$$5y - 80 = 3y \implies 2y = 80 \implies y = 40 \text{ cm}$$

So another point is 40 cm from $q_1$ (i.e., 24 cm from $q_2$, on the far side of $q_2$).

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Answer: The electric potential is zero at two points on the line joining the charges:
1. 10 cm from the positive charge ($5 \times 10^{-8}$ C), between the two charges.
2. 40 cm from the positive charge (24 cm from the negative charge), outside the two charges on the side of the negative charge.

Given:
- Side of regular hexagon: $a = 10$ cm $= 0.1$ m
- Charge at each vertex: $q = 5\,\mu\text{C} = 5 \times 10^{-6}$ C
- Number of vertices: 6

Key Fact: In a regular hexagon, the distance from each vertex to the centre equals the side length. So $r = a = 0.1$ m.

Concept: Potential is a scalar quantity. By the superposition principle:

$$V = 6 \times \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$$

Calculation:

$$V = 6 \times (9 \times 10^9) \times \frac{5 \times 10^{-6}}{0.1}$$

$$V = 6 \times 9 \times 10^9 \times 5 \times 10^{-5}$$

$$V = 6 \times 45 \times 10^4$$

$$\boxed{V = 2.7 \times 10^6 \text{ V}}$$

Answer: The electric potential at the centre of the hexagon is $2.7 \times 10^6$ V.

Given:
- $q_A = +2\,\mu$C at point A
- $q_B = -2\,\mu$C at point B
- Distance AB = 6 cm

This system constitutes an electric dipole.

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(a) Identifying the equipotential surface:

For any point P on the perpendicular bisector of AB, the distances from P to A and from P to B are equal: $r_{PA} = r_{PB} = r$.

The potential at P:
$$V_P = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_A}{r} + \frac{q_B}{r}\right) = \frac{1}{4\pi\varepsilon_0}\left(\frac{+2\,\mu\text{C}}{r} + \frac{-2\,\mu\text{C}}{r}\right) = 0$$

Since $V = 0$ for every point on the perpendicular bisector of AB, the plane perpendicular to AB and passing through its midpoint is an equipotential surface with $V = 0$.

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(b) Direction of electric field on this surface:

The electric field at every point on an equipotential surface is perpendicular to the equipotential surface.

Since the equipotential surface is the perpendicular bisector plane of AB, the electric field at every point on this surface is directed along the line AB (i.e., from the positive charge A towards the negative charge B, which is the direction of the dipole axis).

Answer:
- (a) The plane perpendicular to AB passing through its midpoint (the perpendicular bisector plane) is the equipotential surface with $V = 0$.
- (b) The electric field at every point on this surface is directed normal to the plane (i.e., along the line joining A to B, from $+2\,\mu$C to $-2\,\mu$C).

Given:
- Radius of spherical conductor: $R = 12$ cm $= 0.12$ m
- Charge: $Q = 1.6 \times 10^{-7}$ C

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(a) Electric field inside the sphere:

For a conductor, the electric field inside is zero.

$$\boxed{E_{\text{inside}} = 0}$$

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(b) Electric field just outside the sphere:

Just outside a charged conductor, $E = \dfrac{\sigma}{\varepsilon_0}$, which is equivalent to treating the sphere as a point charge:

$$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$$

$$E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2}$$

$$E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{0.0144}$$

$$E = \frac{1.44 \times 10^3}{0.0144} = 1.0 \times 10^5 \text{ N/C}$$

$$\boxed{E_{\text{just outside}} = 1.0 \times 10^5 \text{ N C}^{-1}}$$

The field is directed radially outward.

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(c) Electric field at a point 18 cm from the centre:

Here $r = 18$ cm $= 0.18$ m > R, so we treat the sphere as a point charge:

$$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$

$$E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2}$$

$$E = \frac{1.44 \times 10^3}{0.0324} = 4.44 \times 10^4 \text{ N/C}$$

$$\boxed{E = 4.4 \times 10^4 \text{ N C}^{-1}}$$

The field is directed radially outward.

Given:
- Initial capacitance (air): $C_0 = 8$ pF
- New distance: $d' = d/2$
- Dielectric constant: $K = 6$

Formula: For a parallel plate capacitor:
$$C_0 = \varepsilon_0 \frac{A}{d}$$

When the distance is halved and a dielectric of constant $K$ is inserted:
$$C' = K \varepsilon_0 \frac{A}{d'} = K \varepsilon_0 \frac{A}{d/2} = 2K \varepsilon_0 \frac{A}{d} = 2K C_0$$

Calculation:
$$C' = 2 \times 6 \times 8 \text{ pF} = 96 \text{ pF}$$

$$\boxed{C' = 96 \text{ pF}}$$

Answer: The new capacitance is 96 pF.

Given:
- $C_1 = C_2 = C_3 = 9$ pF
- Supply voltage: $V = 120$ V
- Connection: Series

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(a) Total capacitance in series:

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}$$

$$\boxed{C = 3 \text{ pF}}$$

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(b) Potential difference across each capacitor:

In a series combination, the same charge $Q$ appears on each capacitor.

$$Q = CV = 3 \times 10^{-12} \times 120 = 3.6 \times 10^{-10} \text{ C}$$

Potential difference across each capacitor:
$$V_1 = V_2 = V_3 = \frac{Q}{C_1} = \frac{3.6 \times 10^{-10}}{9 \times 10^{-12}} = 40 \text{ V}$$

Verification: $V_1 + V_2 + V_3 = 40 + 40 + 40 = 120$ V ✓

$$\boxed{V_{\text{each}} = 40 \text{ V}}$$

Given:
- $C_1 = 2$ pF, $C_2 = 3$ pF, $C_3 = 4$ pF
- Supply voltage: $V = 100$ V
- Connection: Parallel

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(a) Total capacitance in parallel:

$$C = C_1 + C_2 + C_3 = 2 + 3 + 4 = 9 \text{ pF}$$

$$\boxed{C = 9 \text{ pF}}$$

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(b) Charge on each capacitor:

In a parallel combination, the voltage across each capacitor equals the supply voltage $V = 100$ V.

$$Q_1 = C_1 V = 2 \times 10^{-12} \times 100 = 2 \times 10^{-10} \text{ C} = 200 \text{ pC}$$

$$Q_2 = C_2 V = 3 \times 10^{-12} \times 100 = 3 \times 10^{-10} \text{ C} = 300 \text{ pC}$$

$$Q_3 = C_3 V = 4 \times 10^{-12} \times 100 = 4 \times 10^{-10} \text{ C} = 400 \text{ pC}$$

$$\boxed{Q_1 = 200\text{ pC},\quad Q_2 = 300\text{ pC},\quad Q_3 = 400\text{ pC}}$$

Given:
- Area of each plate: $A = 6 \times 10^{-3}$ m²
- Distance between plates: $d = 3$ mm $= 3 \times 10^{-3}$ m
- Voltage: $V = 100$ V
- $\varepsilon_0 = 8.854 \times 10^{-12}$ C² N⁻¹ m⁻²

Formula:
$$C = \varepsilon_0 \frac{A}{d}$$

Calculation of capacitance:
$$C = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}$$

$$C = \frac{8.854 \times 10^{-12} \times 6}{3} = 8.854 \times 10^{-12} \times 2$$

$$\boxed{C = 17.71 \times 10^{-12} \text{ F} \approx 17.71 \text{ pF}}$$

Calculation of charge:
$$Q = CV = 17.71 \times 10^{-12} \times 100$$

$$\boxed{Q = 1.771 \times 10^{-9} \text{ C} \approx 1.77 \text{ nC}}$$

Answer: Capacitance $\approx 17.71$ pF; Charge on each plate $\approx 1.77 \times 10^{-9}$ C.

From Exercise 2.8: $C_0 = 17.71$ pF, $V_0 = 100$ V, $Q_0 = 1.771 \times 10^{-9}$ C

Dielectric constant of mica: $K = 6$

When a dielectric of constant $K$ fills the entire space between the plates, the new capacitance is:
$$C = KC_0 = 6 \times 17.71 \text{ pF} = 106.26 \text{ pF}$$

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(a) While the voltage supply remains connected ($V$ = constant = 100 V):

- The voltage across the capacitor remains fixed at $V = 100$ V.
- The new capacitance: $C = KC_0 = 6 \times 17.71 = 106.26$ pF
- The new charge:
$$Q = CV = 106.26 \times 10^{-12} \times 100 = 1.0626 \times 10^{-8} \text{ C}$$
- The electric field between the plates: $E = V/d = 100/(3\times10^{-3}) = 3.33 \times 10^4$ V/m (unchanged, since $V$ and $d$ are unchanged).

Summary: Capacitance increases 6-fold to ~106.26 pF; charge increases 6-fold to ~$1.06 \times 10^{-8}$ C; voltage remains 100 V; electric field remains the same.

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(b) After the supply is disconnected ($Q$ = constant = $1.771 \times 10^{-9}$ C):

- The charge on the plates remains fixed (no path for charge to flow).
- The new capacitance: $C = KC_0 = 106.26$ pF
- The new voltage:
$$V' = \frac{Q}{C} = \frac{1.771 \times 10^{-9}}{106.26 \times 10^{-12}} = \frac{100}{6} \approx 16.7 \text{ V}$$
- The new electric field:
$$E' = \frac{V'}{d} = \frac{16.7}{3 \times 10^{-3}} \approx 5.56 \times 10^3 \text{ V/m}$$

Summary: Capacitance increases 6-fold to ~106.26 pF; charge remains $1.771 \times 10^{-9}$ C; voltage decreases to ~16.7 V; electric field decreases to ~$5.56 \times 10^3$ V/m.

Given:
- Capacitance: $C = 12$ pF $= 12 \times 10^{-12}$ F
- Voltage: $V = 50$ V

Formula: Energy stored in a capacitor:
$$U = \frac{1}{2}CV^2$$

Calculation:
$$U = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2$$

$$U = \frac{1}{2} \times 12 \times 10^{-12} \times 2500$$

$$U = \frac{1}{2} \times 3 \times 10^{-8}$$

$$\boxed{U = 1.5 \times 10^{-8} \text{ J}}$$

Answer: The electrostatic energy stored in the capacitor is $1.5 \times 10^{-8}$ J.

Given:
- $C_1 = C_2 = 600$ pF $= 600 \times 10^{-12}$ F
- Initial voltage: $V = 200$ V
- $C_2$ is initially uncharged.

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Step 1: Initial energy stored in $C_1$:

$$U_i = \frac{1}{2}C_1 V^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2$$

$$U_i = \frac{1}{2} \times 600 \times 10^{-12} \times 4 \times 10^4 = 1.2 \times 10^{-5} \text{ J}$$

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Step 2: Common voltage after connection:

Initial charge on $C_1$: $Q = C_1 V = 600 \times 10^{-12} \times 200 = 1.2 \times 10^{-7}$ C

When connected to uncharged $C_2$, charge redistributes. By conservation of charge:
$$Q = (C_1 + C_2)V'$$

$$V' = \frac{Q}{C_1 + C_2} = \frac{1.2 \times 10^{-7}}{600 \times 10^{-12} + 600 \times 10^{-12}} = \frac{1.2 \times 10^{-7}}{1.2 \times 10^{-9}} = 100 \text{ V}$$

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Step 3: Final energy stored in both capacitors:

$$U_f = \frac{1}{2}(C_1 + C_2)V'^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2$$

$$U_f = \frac{1}{2} \times 1200 \times 10^{-12} \times 10^4 = 6 \times 10^{-6} \text{ J}$$

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Step 4: Energy lost:

$$\Delta U = U_i - U_f = 1.2 \times 10^{-5} - 6 \times 10^{-6} = 12 \times 10^{-6} - 6 \times 10^{-6}$$

$$\boxed{\Delta U = 6 \times 10^{-6} \text{ J} = 6\,\mu\text{J}}$$

Answer: The electrostatic energy lost in the process is $6 \times 10^{-6}$ J. This energy is dissipated as heat (and electromagnetic radiation) in the connecting wires during the transient current flow.