Electrostatic Potential and Capacitance Class 12 NCERT Solutions & PYQs
🔋 Introduction
Pichle chapter mein humne Electric field dekha. Ab hum baat karenge Electric Potential ki—jo ki ek scalar quantity hai aur problems ko solve karna bahut asaan bana deti hai. Hum padhenge ki kaise energy store hoti hai (Capacitors) aur dielectric materials ka kya impact hota hai. Board exams mein is chapter se capacitors ke combination aur energy storage par pakka questions aate hain.
📐 Chapter Formula Sheet
- Potential due to point charge: \(V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}\)
- Potential Energy: \(U = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r}\)
- Capacitance: \(C = \frac{Q}{V}\)
- Parallel Plate Capacitor: \(C = \frac{\epsilon_0 A}{d}\)
- Energy Stored: \(U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}\)
📚 Part 1: Full NCERT Exercise Solutions
Q 2.1: Two charges \(5 \times 10^{-8}\) C and \(-3 \times 10^{-8}\) C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero?
Ans: Let the point P be at distance \(x\) from the positive charge.
Net Potential \(V_1 + V_2 = 0\)
\(\frac{1}{4\pi\epsilon_0} \left[ \frac{5 \times 10^{-8}}{x} + \frac{-3 \times 10^{-8}}{0.16 - x} \right] = 0\)
\(\frac{5}{x} = \frac{3}{0.16 - x} \Rightarrow 5(0.16 - x) = 3x\)
\(0.80 - 5x = 3x \Rightarrow 8x = 0.80 \Rightarrow x = 0.10 \text{ m} = \mathbf{10 \text{ cm}}\).
(Note: Potential can also be zero at a point outside the charges, which comes to be 40 cm from the positive charge).
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.
Ans: In a regular hexagon, the distance from center to each vertex is equal to the side, \(r = 10 \text{ cm} = 0.1 \text{ m}\).
There are 6 vertices, so total potential \(V = 6 \times \frac{1}{4\pi\epsilon_0} \frac{q}{r}\)
\(V = 6 \times (9 \times 10^9) \times \frac{5 \times 10^{-6}}{0.1}\)
\(V = \mathbf{2.7 \times 10^6 \text{ V}}\).
Q 2.3: Two charges 2 \(\mu\)C and -2 \(\mu\)C are placed at points A and B, 6 cm apart. (a) Identify an equipotential surface. (b) What is the direction of electric field?
Ans: (a) The plane normal to the line AB and passing through its mid-point is the equipotential surface where potential is zero everywhere.
(b) The electric field is always normal to the equipotential surface. Here, it is directed from positive charge A to negative charge B.
Q 2.4: A spherical conductor of radius 12 cm has a charge of \(1.6 \times 10^{-7}\) C distributed uniformly on its surface. What is the electric field (a) inside (b) just outside (c) at 18 cm from centre?
Ans: (a) Inside a conductor, **Electric Field is zero**.
(b) Just outside: \(E = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2} = \mathbf{10^5 \text{ N/C}}\).
(c) At 18 cm: \(E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} = \mathbf{4.44 \times 10^4 \text{ N/C}}\).
Q 2.5: A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance is reduced by half and the space filled with a substance of dielectric constant 6?
Ans: Initial \(C_0 = \frac{\epsilon_0 A}{d} = 8 \text{ pF}\).
New \(C = \frac{K \epsilon_0 A}{d/2} = 2K \left( \frac{\epsilon_0 A}{d} \right) = 2 \times 6 \times 8 = \mathbf{96 \text{ pF}}\).
🔥 Part 2: 5 Most Repeated PYQs (Board Favorites)
Q1: Derive the expression for capacitance of a parallel plate capacitor. What is the effect of dielectric?
Ans: Consider plates of area \(A\) with charge \(+Q\) and \(-Q\) separated by distance \(d\).
Electric field \(E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}\).
Potential difference \(V = Ed = \frac{Qd}{A\epsilon_0}\).
Capacitance \(C = \frac{Q}{V} = \frac{Q}{Qd/A\epsilon_0} = \frac{\epsilon_0 A}{d}\).
**Effect of Dielectric:** If a dielectric of constant \(K\) is filled, capacitance increases by factor \(K\), i.e., \(C' = KC\).
Q2: Derive the expression for energy stored in a capacitor.
Ans: Work done in charging a capacitor by small charge \(dq\): \(dW = V dq = \frac{q}{C} dq\).
Total work \(W = \int_{0}^{Q} \frac{q}{C} dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_0^Q\).
This work is stored as energy \(U = \frac{Q^2}{2C} = \frac{1}{2} CV^2\).
Q3: What are Equipotential Surfaces? Give two properties.
Ans: A surface which has the same electric potential at every point is called an equipotential surface.
1. No work is done in moving a charge between two points on it.
2. Electric field is always perpendicular to the surface.
Q4: Explain the combination of capacitors in series and parallel.
Ans:
**Series:** \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ...\)
**Parallel:** \(C_{eq} = C_1 + C_2 + ...\)
⚡ Part 3: 15 Extra descriptive Practice Questions
- Define 1 Volt.
Ans: 1 Volt is the potential difference between two points when 1 Joule of work is done to move a charge of 1 Coulomb from one point to the other. - Why is potential constant inside a charged hollow sphere?
Ans: The electric field (\(E\)) inside a hollow conductor is zero. Since \(E = -dV/dr = 0\), the potential \(V\) must be a constant value everywhere inside the sphere. - What is a Dielectric? Give examples.
Ans: A dielectric is a non-conducting material (insulator) which transmits electric effect without actually conducting electricity. Examples: Mica, Glass, Water, Wood. - Define Polarization of a dielectric.
Ans: It is the process of inducing equal and opposite charges on the two faces of a dielectric when placed in an external electric field. It is the net dipole moment per unit volume. - What is the unit of Capacitance? Define it.
Ans: The SI unit is Farad (F). 1 Farad is the capacitance of a conductor which requires a charge of 1 Coulomb to raise its potential by 1 Volt. - What is the work done in moving a charge on an equipotential surface?
Ans: Zero. Since the potential difference between any two points on an equipotential surface is zero (\(\Delta V = 0\)), the work done \(W = q \Delta V = 0\). - How does capacitance change if distance \(d\) increases?
Ans: Since capacitance \(C = \frac{\epsilon_0 A}{d}\), it is inversely proportional to distance. So, if \(d\) increases, the capacitance \(C\) decreases. - Can two equipotential surfaces intersect? Why?
Ans: No. If they intersect, there would be two different values of electric potential at the point of intersection, which is physically impossible. - Define Electric Potential Energy.
Ans: The work done by an external force in bringing a system of point charges from infinity to their respective locations to form a system. - What is the potential at infinity?
Ans: By universally accepted convention, the electric potential at an infinite distance from a charge is considered to be zero. - Write the relation between Electric field and Potential gradient.
Ans: \(E = -\frac{dV}{dr}\). Electric field is equal to the negative of the potential gradient. - What is a Polar molecule?
Ans: A molecule in which the centers of positive and negative charges do not coincide even in the absence of an external electric field, resulting in a permanent dipole moment (e.g., \(H_2O\), \(HCl\)). - Define Dielectric Strength.
Ans: The maximum electric field that a dielectric medium can withstand without breakdown (i.e., without ionizing and starting to conduct). - Calculate the potential at the surface of a gold nucleus (\(Z=79, r=6.6 \times 10^{-15} \text{ m}\)).
Ans: \(V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} = 9 \times 10^9 \times \frac{79 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-15}} \approx \mathbf{1.72 \times 10^7 \text{ V}}\). - If \(V\) is constant, what is the value of \(E\)?
Ans: If \(V\) is constant, the change in potential \(dV = 0\). Using \(E = -dV/dr\), the electric field \(E\) will be zero.
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