Rbse Solution for Class 12 Physics Chapter 2 : Electrostatic Potential and Capacitance
| Chapter 2 | (Open) |
2.1 Two charges $q_1 = 5 \times 10^{-8}\text{ C}$ and $q_2 = -3 \times 10^{-8}\text{ C}$ are located $16\text{ cm}$ 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.
Let the charges be placed on the $x$-axis, with $q_1$ at the origin $(x=0)$ and $q_2$ at $x=d = 0.16\text{ m}$. The potential $V$ at a point $P$ due to these two charges is the algebraic sum of the potentials due to $q_1$ and $q_2$.
$$V = V_1 + V_2 = k \frac{q_1}{r_1} + k \frac{q_2}{r_2}$$
We require $V=0$, so:
$$k \frac{q_1}{r_1} + k \frac{q_2}{r_2} = 0 \implies \frac{q_1}{r_1} = – \frac{q_2}{r_2}$$
$$\frac{5 \times 10^{-8}}{r_1} = – \frac{-3 \times 10^{-8}}{r_2} \implies \frac{5}{r_1} = \frac{3}{r_2}$$
$$5 r_2 = 3 r_1 \implies r_1 = \frac{5}{3} r_2 \quad \dots(1)$$
There are two regions where the potential can be zero:
Case 1: Point $P$ lies between the charges.
If $P$ is at distance $x$ from $q_1$, then $r_1 = x$ and $r_2 = d – x = 0.16 – x$.
Substituting into (1):
$$x = \frac{5}{3} (0.16 – x)$$
$$3x = 5(0.16) – 5x$$
$$8x = 0.80$$
$$x = \frac{0.80}{8} = 0.10\text{ m} = 10\text{ cm}$$
Point 1: $10\text{ cm}$ from the $5 \times 10^{-8}\text{ C}$ charge and $6\text{ cm}$ from the $-3 \times 10^{-8}\text{ C}$ charge.
Case 2: Point $P$ lies outside the charges, on the side of the smaller magnitude charge ($q_2$).
Let $P$ be at distance $y$ from $q_2$ (and $d+y$ from $q_1$). Then $r_2 = y$ and $r_1 = d + y = 0.16 + y$.
Substituting into (1):
$$0.16 + y = \frac{5}{3} y$$
$$3(0.16 + y) = 5y$$
$$0.48 + 3y = 5y$$
$$2y = 0.48$$
$$y = 0.24\text{ m} = 24\text{ cm}$$
Point 2: $24\text{ cm}$ from the $-3 \times 10^{-8}\text{ C}$ charge and $40\text{ cm}$ from the $5 \times 10^{-8}\text{ C}$ charge.
2.2 A regular hexagon of side $10\text{ cm}$ has a charge $5\text{ $\mu$C}$ at each of its vertices. Calculate the potential at the centre of the hexagon.
- Side of hexagon $s = 10\text{ cm} = 0.10\text{ m}$.
- Charge at each vertex $q = 5\text{ $\mu$C} = 5 \times 10^{-6}\text{ C}$.
- Number of charges $N = 6$.
In a regular hexagon, the distance $r$ from the centre to each vertex is equal to the side length $s$.
$$r = s = 0.10\text{ m}$$
The electric potential $V$ at the centre is the algebraic sum of the potentials due to all six charges:
$$V = \sum_{i=1}^{6} V_i = \sum_{i=1}^{6} k \frac{q_i}{r}$$
Since all charges $q_i$ are identical ($q$) and the distances $r$ are the same:
$$V = 6 \times \left( k \frac{q}{r} \right)$$
$$V = 6 \times (9 \times 10^9\text{ N m}^2\text{ C}^{-2}) \frac{5 \times 10^{-6}\text{ C}}{0.10\text{ m}}$$
$$V = 6 \times (9 \times 10^9) \times (50 \times 10^{-6})\text{ V}$$
$$V = 2700 \times 10^3\text{ V} = 2.7 \times 10^6\text{ V}$$
The potential at the centre of the hexagon is $2.7 \times 10^6\text{ V}$.
2.3 Two charges $q_A = 2\text{ $\mu$C}$ and $q_B = -2\text{ $\mu$C}$ are placed at points A and B $6\text{ cm}$ apart.
(a) Identify an equipotential surface of the system.
The system is an electric dipole (two equal and opposite charges separated by a distance).
An equipotential surface is a surface on which the electric potential $V$ is zero. For a dipole, the plane which is perpendicular to the dipole axis (line AB) and passes through the midpoint $O$ of the dipole axis is an equipotential surface where $V=0$. This is because for any point $P$ on this plane, the distance $r_A$ from $q_A$ is equal to the distance $r_B$ from $q_B$.
$$V_P = k \frac{q_A}{r_A} + k \frac{q_B}{r_B} = k \frac{2\text{ $\mu$C}}{r} + k \frac{-2\text{ $\mu$C}}{r} = 0$$
The equipotential surface is the plane normal to the line AB and passing through the midpoint of AB.
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(b) What is the direction of the electric field at every point on this surface?
The electric field $\vec{E}$ is always perpendicular to the equipotential surface at every point.
For the $V=0$ equipotential plane of the dipole, the electric field is directed parallel to the dipole axis (line AB), pointing from the positive charge ($q_A$) towards the negative charge ($q_B$).
The direction of the electric field is perpendicular to the surface and parallel to the line AB (from A to B).
2.4 A spherical conductor of radius $R = 12\text{ cm}$ has a charge of $Q = 1.6 \times 10^{-7}\text{ C}$ distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point $18\text{ cm}$ from the centre of the sphere?
A charged spherical conductor is an equipotential body. For external points, the charge $Q$ can be considered to be concentrated at the centre.
- $R = 0.12\text{ m}$
- $Q = 1.6 \times 10^{-7}\text{ C}$
(a) inside the sphere ($r < R$)
For a charged conductor, the net electric field inside the body is zero due to the free movement of charges that redistribute until equilibrium is reached.
$$E_{\text{inside}} = 0$$
(b) just outside the sphere ($r = R$)
$$E_{\text{outside}} = k \frac{Q}{R^2}$$
$$E_{\text{outside}} = (9 \times 10^9\text{ N m}^2\text{ C}^{-2}) \frac{1.6 \times 10^{-7}\text{ C}}{(0.12\text{ m})^2}$$
$$E_{\text{outside}} = \frac{14.4 \times 10^2}{0.0144}\text{ N/C} = 1000 \times 10^2\text{ N/C} = 1.0 \times 10^5\text{ N/C}$$
The electric field just outside is $1.0 \times 10^5\text{ N/C}$ (radially outward).
(c) at a point $18\text{ cm}$ from the centre of the sphere ($r = 0.18\text{ m}$)
$$E_{\text{point}} = k \frac{Q}{r^2}$$
$$E_{\text{point}} = (9 \times 10^9\text{ N m}^2\text{ C}^{-2}) \frac{1.6 \times 10^{-7}\text{ C}}{(0.18\text{ m})^2}$$
$$E_{\text{point}} = \frac{14.4 \times 10^2}{0.0324}\text{ N/C} \approx 444.4 \times 10^2\text{ N/C} \approx 4.44 \times 10^4\text{ N/C}$$
The electric field at $18\text{ cm}$ is $4.44 \times 10^4\text{ N/C}$ (radially outward).
2.5 A parallel plate capacitor with air between the plates has a capacitance of $C_0 = 8\text{ pF}$ ($1\text{ pF} = 10^{-12}\text{ F}$). 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 $K=6$?
The capacitance $C_0$ of a parallel plate capacitor with air (or vacuum, $K_0=1$) is:
$$C_0 = \frac{\epsilon_0 A}{d}$$
The new capacitance $C$ is given by:
$$C = \frac{K \epsilon_0 A}{d’}$$
where $K=6$ is the new dielectric constant and $d’$ is the new distance.
The problem states:
- Distance is reduced by half: $d’ = d/2$.
- Space is filled with $K=6$.
Substitute these into the formula for $C$:
$$C = \frac{K \epsilon_0 A}{d/2} = K \cdot 2 \cdot \left( \frac{\epsilon_0 A}{d} \right)$$
$$C = 2 K C_0$$
$$C = 2 \times 6 \times C_0 = 12 C_0$$
$$C = 12 \times 8\text{ pF} = 96\text{ pF}$$
The new capacitance will be $96\text{ pF}$.
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2.6 Three capacitors each of capacitance $C = 9\text{ pF}$ are connected in series.
(a) What is the total capacitance of the combination?
For capacitors connected in series, the reciprocal of the equivalent capacitance $C_S$ is the sum of the reciprocals of individual capacitances:
$$\frac{1}{C_S} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$
Since $C_1 = C_2 = C_3 = 9\text{ pF}$:
$$\frac{1}{C_S} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}\text{ pF}^{-1}$$
$$C_S = 3\text{ pF}$$
The total capacitance is $3\text{ pF}$.
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(b) What is the potential difference across each capacitor if the combination is connected to a $120\text{ V}$ supply?
In a series combination, the charge $Q$ is the same across all capacitors.
The total charge stored is $Q = C_S V_{\text{supply}}$.
$$Q = (3 \times 10^{-12}\text{ F}) \times (120\text{ V}) = 360 \times 10^{-12}\text{ C}$$
The potential difference $V$ across each capacitor $C$ is $V = Q/C$.
Since all capacitors have the same capacitance ($C_1=C_2=C_3$), the potential difference across each will be equal:
$$V_1 = V_2 = V_3 = \frac{V_{\text{supply}}}{3} = \frac{120\text{ V}}{3} = 40\text{ V}$$
The potential difference across each capacitor is $40\text{ V}$.
2.7 Three capacitors of capacitances $C_1 = 2\text{ pF}$, $C_2 = 3\text{ pF}$ and $C_3 = 4\text{ pF}$ are connected in parallel.
(a) What is the total capacitance of the combination?
For capacitors connected in parallel, the total equivalent capacitance $C_P$ is the sum of the individual capacitances:
$$C_P = C_1 + C_2 + C_3$$
$$C_P = 2\text{ pF} + 3\text{ pF} + 4\text{ pF} = 9\text{ pF}$$
The total capacitance is $9\text{ pF}$.
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(b) Determine the charge on each capacitor if the combination is connected to a $100\text{ V}$ supply.
In a parallel combination, the potential difference $V$ is the same across all capacitors: $V = 100\text{ V}$.
The charge $Q$ on each capacitor is given by $Q = C V$.
$$Q_1 = C_1 V = (2 \times 10^{-12}\text{ F}) \times (100\text{ V}) = 200 \times 10^{-12}\text{ C} = 2.0 \times 10^{-10}\text{ C}$$
$$Q_2 = C_2 V = (3 \times 10^{-12}\text{ F}) \times (100\text{ V}) = 300 \times 10^{-12}\text{ C} = 3.0 \times 10^{-10}\text{ C}$$
$$Q_3 = C_3 V = (4 \times 10^{-12}\text{ F}) \times (100\text{ V}) = 400 \times 10^{-12}\text{ C} = 4.0 \times 10^{-10}\text{ C}$$
The charges are: $Q_1 = 2.0 \times 10^{-10}\text{ C}$, $Q_2 = 3.0 \times 10^{-10}\text{ C}$, and $Q_3 = 4.0 \times 10^{-10}\text{ C}$.
2.8 In a parallel plate capacitor with air between the plates, each plate has an area $A = 6 \times 10^{-3}\text{ m}^2$ and the distance between the plates is $d = 3\text{ mm}$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100\text{ V}$ supply, what is the charge on each plate of the capacitor?
- $A = 6 \times 10^{-3}\text{ m}^2$
- $d = 3\text{ mm} = 3 \times 10^{-3}\text{ m}$
- $V = 100\text{ V}$
- $\epsilon_0 \approx 8.85 \times 10^{-12}\text{ F/m}$
Capacitance $C_0$
$$C_0 = \frac{\epsilon_0 A}{d}$$
$$C_0 = \frac{(8.85 \times 10^{-12}\text{ F/m}) (6 \times 10^{-3}\text{ m}^2)}{3 \times 10^{-3}\text{ m}}$$
$$C_0 = 8.85 \times 10^{-12} \times 2\text{ F} = 17.7 \times 10^{-12}\text{ F} = 17.7\text{ pF}$$
The capacitance is $17.7\text{ pF}$.
Charge $Q$
$$Q = C_0 V$$
$$Q = (17.7 \times 10^{-12}\text{ F}) \times (100\text{ V})$$
$$Q = 1770 \times 10^{-12}\text{ C} = 1.77 \times 10^{-9}\text{ C}$$
The charge on each plate is $1.77 \times 10^{-9}\text{ C}$.
2.9 Explain what would happen if in the capacitor given in Exercise 2.8 ($C_0 = 17.7\text{ pF}$, $d=3\text{ mm}$), a $3\text{ mm}$ thick mica sheet (dielectric constant $K=6$) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.
The mica sheet fills the entire space ($t=d$). The new capacitance $C$ is:
$$C = K C_0 = 6 \times 17.7\text{ pF} = 106.2\text{ pF}$$
(a) While the voltage supply remained connected ($V$ is constant)
- Capacitance ($C$): Increases by a factor of $K=6$ to $106.2\text{ pF}$.
- Potential Difference ($V$): Remains constant at $100\text{ V}$ because it is connected to the supply.
- Charge ($Q = C V$): Increases by a factor of $K=6$. $Q_{\text{new}} = 6 \times Q_0 = 6 \times (1.77 \times 10^{-9}\text{ C}) = 1.062 \times 10^{-8}\text{ C}$. The extra charge is supplied by the battery.
- Energy ($U = \frac{1}{2} C V^2$): Increases by a factor of $K=6$.
(b) After the supply was disconnected ($Q$ is constant)
- Capacitance ($C$): Increases by a factor of $K=6$ to $106.2\text{ pF}$.
- Charge ($Q$): Remains constant at $1.77 \times 10^{-9}\text{ C}$ because there is no path for the charge to leave the isolated plates.
- Potential Difference ($V = Q/C$): Decreases by a factor of $K=6$.$$V_{\text{new}} = \frac{V_0}{K} = \frac{100\text{ V}}{6} \approx 16.7\text{ V}$$
- Energy ($U = \frac{Q^2}{2C}$): Decreases by a factor of $K=6$. The lost energy is dissipated as heat or used to pull the dielectric into the capacitor.
2.10 A $12\text{ pF}$ capacitor is connected to a $50\text{ V}$ battery. How much electrostatic energy is stored in the capacitor?
- $C = 12\text{ pF} = 12 \times 10^{-12}\text{ F}$
- $V = 50\text{ V}$
The electrostatic potential energy $U$ stored in a capacitor is:
$$U = \frac{1}{2} C V^2$$
$$U = \frac{1}{2} (12 \times 10^{-12}\text{ F}) (50\text{ V})^2$$
$$U = (6 \times 10^{-12}\text{ F}) \times (2500\text{ V}^2)$$
$$U = 15000 \times 10^{-12}\text{ J} = 1.5 \times 10^{-8}\text{ J}$$
The electrostatic energy stored is $1.5 \times 10^{-8}\text{ J}$.
2.11 A $C_1 = 600\text{ pF}$ capacitor is charged by a $V_0 = 200\text{ V}$ supply. It is then disconnected from the supply and is connected to another uncharged $C_2 = 600\text{ pF}$ capacitor. How much electrostatic energy is lost in the process?
Step 1: Energy Stored in $C_1$ Initially
Initial charge on $C_1$:
$$Q_1 = C_1 V_0 = (600 \times 10^{-12}\text{ F}) \times (200\text{ V}) = 1.2 \times 10^{-7}\text{ C}$$
Initial energy stored in the system ($C_1$ alone):
$$U_{\text{initial}} = \frac{1}{2} C_1 V_0^2 = \frac{1}{2} (600 \times 10^{-12}\text{ F}) (200\text{ V})^2$$
$$U_{\text{initial}} = (300 \times 10^{-12}) \times (40000)\text{ J} = 12 \times 10^{-6}\text{ J}$$
Step 2: Connection and Final State
When $C_1$ (charged) is connected to $C_2$ (uncharged) in parallel, the total charge $Q_1$ is conserved and is redistributed between $C_1$ and $C_2$. The capacitors share a common final potential $V_f$.
Since $C_1 = C_2 = C$, the final charge on each is $Q_f = Q_1 / 2$.
The equivalent capacitance is $C_{\text{eq}} = C_1 + C_2 = 1200\text{ pF}$.
The final common potential is:
$$V_f = \frac{Q_1}{C_{\text{eq}}} = \frac{1.2 \times 10^{-7}\text{ C}}{1200 \times 10^{-12}\text{ F}} = 100\text{ V}$$
Final energy stored in the system:
$$U_{\text{final}} = \frac{1}{2} C_{\text{eq}} V_f^2 = \frac{1}{2} (1200 \times 10^{-12}\text{ F}) (100\text{ V})^2$$
$$U_{\text{final}} = (600 \times 10^{-12}) \times (10000)\text{ J} = 6 \times 10^{-6}\text{ J}$$
Step 3: Energy Lost
The energy lost ($\Delta U$) is the difference between the initial and final stored energies. This energy is lost primarily as heat during the charge redistribution process (sparking/current flow).
$$\Delta U = U_{\text{initial}} – U_{\text{final}}$$
$$\Delta U = 12 \times 10^{-6}\text{ J} – 6 \times 10^{-6}\text{ J} = 6 \times 10^{-6}\text{ J}$$
The electrostatic energy lost in the process is $6 \times 10^{-6}\text{ J}$ (or $6\text{ $\mu$J}$).
