Potential $V = kq/r$. For zero potential, contributions from all charges add to zero. For a conductor, $E = 0$ inside and $E = \sigma/\varepsilon_0$ just outside the surface.
Q 2.1
Two charges $5\times10^{-8}$ C and $-3\times10^{-8}$ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
$q_1 = 5\times10^{-8}$ C at origin O, $q_2 = -3\times10^{-8}$ C at 16 cm from O
P₁ — internal zero potential point; P₂ — external zero potential point
Internal Point P₁
①
Let P₁ be at distance $x$ from $q_1$, so distance from $q_2$ is $(16-x)$ cm. For $V = 0$:
$$\frac{kq_1}{x} + \frac{kq_2}{16-x} = 0 \implies \frac{5}{x} = \frac{3}{16-x}$$
②
$$5(16-x) = 3x \implies 80 = 8x$$
★ P₁
$$x = 10 \text{ cm from } q_1 \quad \text{(between the charges)}$$
External Point P₂
①
Let P₂ be at distance $x$ from $q_1$, beyond $q_2$. Distance from $q_2$ is $(x-16)$ cm:
$$\frac{5}{x} = \frac{3}{x-16} \implies 5(x-16) = 3x \implies 2x = 80$$
★ P₂
$$x = 40 \text{ cm from } q_1 \quad \text{(i.e. 24 cm beyond }q_2\text{)}$$
Q 2.2
A regular hexagon of side 10 cm has a charge $5\ \mu$C at each of its vertices. Calculate the potential at the centre of the hexagon.
Side $= r = 10$ cm $= 0.1$ m, $q = 5\times10^{-6}$ C at each of 6 vertices
All 6 vertices are equidistant from centre O — distance = side = 10 cm
①
For a regular hexagon, each vertex is at distance $r = $ side $= 0.1$ m from the centre. Total potential:
$$V = 6\times\frac{kq}{r}$$
Two charges $2\ \mu$C and $-2\ \mu$C are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?
$q_A = +2\,\mu$C, $q_B = -2\,\mu$C, $AB = 6$ cm — an electric dipole
Part (a) — Equipotential Surface
①
Since the charges are equal and opposite, the zero-potential surface is the plane that perpendicularly bisects the line segment AB. At any point P on this plane, $r_{PA} = r_{PB}$, so:
$$V = \frac{kq_A}{r_{PA}} + \frac{kq_B}{r_{PB}} = \frac{k(+2\mu C)}{r} + \frac{k(-2\mu C)}{r} = 0$$
★ (a)
The $V = 0$ equipotential surface is the plane passing through the midpoint of AB and perpendicular to AB.
Part (b) — Direction of E on the surface
①
Electric field is always perpendicular to an equipotential surface. On the perpendicular bisector plane, by symmetry, the resultant field points along the direction from $A\ (+)$ to $B\ (-)$, i.e., normal to this plane — from the positive charge to the negative charge.
★ (b)
The field $\vec{E}$ is perpendicular to the equipotential surface, directed from $A\ (+2\,\mu\text{C})$ toward $B\ (-2\,\mu\text{C})$.
Q 2.4
A spherical conductor of radius 12 cm has a charge of $1.6\times10^{-7}$ C distributed uniformly on its surface. What is the electric field (a) inside the sphere, (b) just outside the sphere, (c) at a point 18 cm from the centre?
$R = 0.12$ m, $q = 1.6\times10^{-7}$ C
E is zero inside; field emerges radially outward from the surface
Part (a) — Inside
★ (a)
$$E_{\text{inside}} = 0 \quad \text{(conductor: all charge resides on surface)}$$
A parallel plate capacitor with air between the plates has a capacitance of 8 pF ($1\text{ pF} = 10^{-12}$ F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
$C_1 = C_2 = C_3 = 9$ pF, series connection, $V = 120$ V
Three 9 pF capacitors in series — same charge on each, voltage divides equally
In series, same charge $Q = CV = 3\times10^{-12}\times120 = 3.6\times10^{-10}$ C on each. So $V_{\text{each}} = Q/C_i$:
$$V_{\text{each}} = \frac{120}{3} = 40 \text{ V}$$
★ (b)
$$V_{\text{across each}} = 40 \text{ V}$$
Q 2.7
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
In parallel, same $V = 100$ V across each:
$$Q_1 = C_1 V = 2\times10^{-12}\times100 = 2\times10^{-10} \text{ C} = 200 \text{ pC}$$
$$Q_2 = 3\times10^{-12}\times100 = 3\times10^{-10} \text{ C} = 300 \text{ pC}$$
$$Q_3 = 4\times10^{-12}\times100 = 4\times10^{-10} \text{ C} = 400 \text{ pC}$$
★ (b)
$Q_1 = 200$ pC, $Q_2 = 300$ pC, $Q_3 = 400$ pC
Q 2.8
In a parallel plate capacitor with air between the plates, each plate has an area of $6\times10^{-3}$ m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate?
$A = 6\times10^{-3}$ m², $d = 3\times10^{-3}$ m, $V = 100$ V
Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant $= 6$) were inserted between the plates, (a) while the voltage supply remained connected, (b) after the supply was disconnected.
Same capacitor: $C_0 = 17.7$ pF, mica $\kappa = 6$, $d_{\text{mica}} = d = 3$ mm. New $C' = \kappa C_0 = 6\times17.7 = 106.2$ pF
Part (a) — Voltage supply remains connected
①
Voltage across plates stays fixed at $V = 100$ V (battery holds it). Capacitance increases to $C' = 106.2$ pF. New charge:
$$Q' = C'V = 106.2\times10^{-12}\times100 = 1.062\times10^{-8} \text{ C}$$
★ (a)
$V$ unchanged at 100 V. Capacitance increases to 106.2 pF. Charge increases from 1.77 nC to 10.62 nC — extra charge flows from battery.
Part (b) — Supply disconnected first
①
Charge $Q = 1.77$ nC is now fixed (no path for charge to change). Capacitance increases to $C' = 106.2$ pF. New voltage:
$$V' = \frac{Q}{C'} = \frac{1.771\times10^{-9}}{106.2\times10^{-12}} = \frac{1.771}{0.1062}\approx 16.7 \text{ V}$$
★ (b)
$Q$ unchanged. Capacitance rises to 106.2 pF. Voltage drops from 100 V to $\approx 16.7$ V.
Q 2.10
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
$\mathcal{E} = 12$ V, $r = 0.4\ \Omega$
①
Maximum current is drawn when the external resistance is zero (short circuit):
$$I_{\max} = \frac{\mathcal{E}}{r} = \frac{12}{0.4}$$
★ Answer
$$I_{\max} = 30 \text{ A}$$
Q 3.2
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is $1.70\times10^{-4}\ ^\circ$C$^{-1}$.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0\times10^{-7}$ m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
$R_1 = 2.1\ \Omega$ at $T_1 = 27.5^\circ$C, $R_2 = 2.7\ \Omega$ at $T_2 = 100^\circ$C
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70\times10^{-4}\ ^\circ$C$^{-1}$.
$V = 230$ V, $I_0 = 3.2$ A (at $27^\circ$C), $I_T = 2.8$ A (steady), $\alpha = 1.70\times10^{-4}\ ^\circ\text{C}^{-1}$
KCL: $\sum I_{\text{in}} = \sum I_{\text{out}}$ at each node. KVL: $\sum \mathcal{E} = \sum IR$ around any loop. Drift velocity: $v_d = I/(nAe)$, time $t = L/v_d$.
Q 3.7
Determine the current in each branch of the network shown in Fig. 3.20.
5 resistors: B–A = 10 Ω, B–C = 5 Ω, A–C = 5 Ω (bridge), A–D = 5 Ω, C–D = 10 Ω. External: 10 Ω in series with 10 V battery (B to D).
Fig 3.20 — Bridge network. Nodes B, A, C, D. Battery 10V + external 10Ω between B and D.
Method: Nodal analysis. Set $V_D = 0$, $V_B = V$ (unknown). Solve KCL at nodes A and C, then find V from the battery equation.
①
Let $V_B = V$, $V_A$ and $V_C$ unknown. At node A (KCL):
$$\frac{V-V_A}{10} = \frac{V_A - V_C}{5} + \frac{V_A}{5}$$
Solving $\Rightarrow V = 5V_A - 2V_C \quad \cdots(1)$
②
At node C (KCL):
$$\frac{V-V_C}{5} + \frac{V_A-V_C}{5} = \frac{V_C}{10}$$
Solving $\Rightarrow V_C = \dfrac{2V+2V_A}{5} \quad \cdots(2)$
③
Substituting (2) into (1): $\;V_A = \dfrac{3V}{7}$, $\;V_C = \dfrac{4V}{7}$.
Total current $I = \dfrac{V_B - V_A}{10} + \dfrac{V_B - V_C}{5} = \dfrac{V}{7}$.
Battery equation: $V = 10 - 10I = 10 - \dfrac{10V}{7}$
$$\Rightarrow 17V = 70 \Rightarrow V = \frac{70}{17},\quad I = \frac{10}{17} \text{ A}$$
★ Branch Currents
$I_{\text{BA}} = \dfrac{4}{17}$ A (through 10 Ω, B→A)
$I_{\text{BC}} = \dfrac{6}{17}$ A (through 5 Ω, B→C)
$I_{\text{CA}} = \dfrac{2}{17}$ A (through 5 Ω bridge, C→A)
$I_{\text{AD}} = \dfrac{6}{17}$ A (through 5 Ω, A→D)
$I_{\text{CD}} = \dfrac{4}{17}$ A (through 10 Ω, C→D)
$I_{\text{total}} = \dfrac{10}{17}$ A $\approx 0.59$ A
Q 3.8
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
$\mathcal{E}_{\text{batt}} = 8$ V (battery being charged), $r = 0.5\ \Omega$, $V_{\text{supply}} = 120$ V, $R_s = 15.5\ \Omega$
①
During charging, EMF of battery opposes current. Total resistance in circuit:
$$I = \frac{V_{\text{supply}} - \mathcal{E}_{\text{batt}}}{R_s + r} = \frac{120 - 8}{15.5 + 0.5} = \frac{112}{16} = 7 \text{ A}$$
②
During charging, terminal voltage of battery (current forced in against EMF):
$$V_T = \mathcal{E} + Ir = 8 + 7\times0.5 = 8 + 3.5$$
★ Terminal V
$$V_T = 11.5 \text{ V}$$
Purpose of series resistor: To limit the charging current. Without it, the current $= (120-8)/0.5 = 224$ A — dangerously large and would damage the battery. The resistor reduces this to a safe 7 A.
Q 3.9
The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5\times10^{28}$ m$^{-3}$. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is $2.0\times10^{-6}$ m² and it is carrying a current of 3.0 A.
$n = 8.5\times10^{28}$ m$^{-3}$, $L = 3.0$ m, $A = 2.0\times10^{-6}$ m², $I = 3.0$ A, $e = 1.6\times10^{-19}$ C
Insight: Individual electrons drift extremely slowly — yet the electrical signal (field) propagates at near the speed of light. It is the wave, not the electrons, that travels fast!
