Question 1

Two charges $5 	imes 10^{-8} \mathrm{C}$ and $-3 	imes 10^{-8} \mathrm{C}$ are located $16 \mathrm{~cm}$ 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.

Accepted Answer

Let the two charges be q1 = 5 × 10⁻⁸ C and q2 = -3 × 10⁻⁸ C, separated by distance d = 16 cm = 0.16 m.

Let the point where potential is zero be at a distance x from q1 along the line joining the charges.

Potential at that point due to q1: V1 = k * q1 / x
Potential at that point due to q2: V2 = k * q2 / (d - x)

Total potential V = V1 + V2 = 0
=> k * q1 / x + k * q2 / (d - x) = 0
=> q1 / x = - q2 / (d - x)

Substitute values:
5 × 10⁻⁸ / x = 3 × 10⁻⁸ / (0.16 - x)

Cross-multiplied:
5 × 10⁻⁸ (0.16 - x) = 3 × 10⁻⁸ x

5 × 10⁻⁸ × 0.16 - 5 × 10⁻⁸ x = 3 × 10⁻⁸ x

(5 × 10⁻⁸ × 0.16) = 3 × 10⁻⁸ x + 5 × 10⁻⁸ x = 8 × 10⁻⁸ x

Calculate left side:
5 × 0.16 = 0.8
So, 0.8 × 10⁻⁸ = 8 × 10⁻⁸ x

Divide both sides by 8 × 10⁻⁸:
0.8 × 10⁻⁸ / 8 × 10⁻⁸ = x
0.1 = x (in meters)

So, x = 0.1 m = 10 cm from q1.

Check if the point lies between the charges or outside:
Since x = 10 cm < 16 cm, the point lies between the charges.

Also check if there is a point outside the charges where potential is zero.

Try point on left side of q1 (x negative):
Potential due to q1 positive, q2 negative, but since q1 > |q2|, potential cannot be zero on left side.

Try point on right side of q2 (x > 16 cm):
Let distance from q2 be y, total distance from q1 is (16 + y).

Potential zero condition:
k * q1 / (16 + y) + k * q2 / y = 0
=> 5 × 10⁻⁸ / (0.16 + y) = 3 × 10⁻⁸ / y

Cross multiply:
5 × 10⁻⁸ y = 3 × 10⁻⁸ (0.16 + y)

5 y = 3 (0.16 + y)

5 y = 0.48 + 3 y

5 y - 3 y = 0.48

2 y = 0.48

y = 0.24 m = 24 cm

So, the second point is 24 cm to the right of q2, i.e., at 16 + 24 = 40 cm from q1.

Therefore, the electric potential is zero at two points on the line joining the charges:
- 10 cm from q1 (between the charges)
- 40 cm from q1 (outside, on the side of the negative charge)

— The potential at a point due to a point charge is V = kq/r. For two charges, total potential is sum of potentials. Setting total potential zero and solving for position gives two points where potential is zero. One lies between the charges, the other outside on the side of the negative charge.

Question 2

A regular hexagon of side $10\mathrm{cm}$ has a charge $5\mu \mathrm{C}$ at each of its vertices. Calculate the potential at the centre of the hexagon.

Accepted Answer

Given: Side of regular hexagon, a = 10 cm = 0.1 m
Charge at each vertex, q = 5 μC = 5 × 10⁻⁶ C

In a regular hexagon, the distance from the center to each vertex (radius) is equal to the side length, r = a = 0.1 m.

Potential at the center due to one charge:
V = k * q / r
where k = 9 × 10⁹ Nm²/C²

Potential at center due to all six charges:
V_total = 6 * V = 6 * (9 × 10⁹) * (5 × 10⁻⁶) / 0.1

Calculate:
V_total = 6 * 9 × 10⁹ * 5 × 10⁻⁶ / 0.1
= 6 * 9 × 10⁹ * 5 × 10⁻⁵
= 6 * 9 * 5 × 10⁴
= 270 × 10⁴
= 2.7 × 10⁶ V

Therefore, the potential at the center of the hexagon is 2.7 × 10⁶ volts. — Potential due to point charges adds algebraically. Since all charges are equal and equidistant from the center, total potential is six times the potential due to one charge at distance equal to side length.

Question 3

Two charges $2\mu \mathrm{C}$ and $-2\mu \mathrm{C}$ 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?

Accepted Answer

(a) The equipotential surfaces of a dipole (two equal and opposite charges) are surfaces where the potential is constant. One such equipotential surface is the plane perpendicular to the line joining the charges and passing through the midpoint between A and B.

(b) The electric field at every point on an equipotential surface is always perpendicular to that surface. Therefore, at every point on this equipotential surface, the electric field direction is normal (perpendicular) to the plane.

Explanation:
Equipotential surfaces are always perpendicular to electric field lines. For a dipole, the plane equidistant from both charges where potential is zero is an equipotential surface. The electric field lines cross this plane perpendicularly. — Equipotential surfaces have constant potential; electric field is always perpendicular to equipotential surfaces. For a dipole, the plane midway between charges is an equipotential surface.

Question 4

A spherical conductor of radius $12\mathrm{cm}$ has a charge of $1.6	imes 10^{-7}\mathrm{C}$ distributed uniformly on its surface. What is the electric field

(a) inside the sphere

(b) just outside the sphere

(c) at a point $18\mathrm{cm}$ from the centre of the sphere?

Accepted Answer

(a) Inside the spherical conductor, the electric field is zero.

(b) Just outside the sphere, the electric field is given by:
E = k * Q / r²
where Q = 1.6 × 10⁻⁷ C, r = radius = 12 cm = 0.12 m, k = 9 × 10⁹ Nm²/C²

E = (9 × 10⁹) * (1.6 × 10⁻⁷) / (0.12)²
= (9 × 10⁹) * (1.6 × 10⁻⁷) / 0.0144
= (1.44 × 10³) / 0.0144
= 1 × 10⁵ N/C

(c) At a point 18 cm = 0.18 m from the center (outside the sphere),

E = k * Q / r²
= (9 × 10⁹) * (1.6 × 10⁻⁷) / (0.18)²
= (1.44 × 10³) / 0.0324
≈ 4.44 × 10⁴ N/C

Therefore,
(a) E = 0 inside the sphere
(b) E = 1 × 10⁵ N/C just outside the sphere
(c) E ≈ 4.44 × 10⁴ N/C at 18 cm from center. — Electric field inside a conductor is zero. Outside, the sphere behaves like a point charge at center. Use Coulomb's law to calculate E at given distances.

Question 5

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 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?

Accepted Answer

Given:
Initial capacitance, C₁ = 8 pF = 8 × 10⁻¹² F
Distance between plates reduced by half: d₂ = d₁ / 2
Dielectric constant, K = 6

Capacitance of parallel plate capacitor:
C = K * ε₀ * A / d

Since C₁ = ε₀ * A / d₁ (air dielectric, K=1)

New capacitance:
C₂ = K * ε₀ * A / d₂ = K * ε₀ * A / (d₁ / 2) = 2 * K * (ε₀ * A / d₁) = 2 * K * C₁

Substitute values:
C₂ = 2 * 6 * 8 pF = 96 pF

Therefore, the new capacitance is 96 pF. — Capacitance is inversely proportional to distance and directly proportional to dielectric constant. Halving distance doubles capacitance; inserting dielectric multiplies capacitance by K.

Question 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 $120\mathrm{V}$ supply?

Accepted Answer

(a) For capacitors in series, total capacitance C_total is given by:
1 / C_total = 1 / C₁ + 1 / C₂ + 1 / C₃
Given C₁ = C₂ = C₃ = 9 pF

1 / C_total = 1/9 + 1/9 + 1/9 = 3/9 = 1/3

=> C_total = 3 pF

(b) Total voltage V = 120 V

Charge Q on each capacitor is same in series:
Q = C_total * V = 3 × 10⁻¹² F * 120 V = 3.6 × 10⁻¹⁰ C

Potential difference across each capacitor:
V_i = Q / C_i = (3.6 × 10⁻¹⁰) / (9 × 10⁻¹²) = 40 V

Therefore, each capacitor has 40 V across it. — In series, reciprocal capacitances add. Charge is same on each capacitor; voltage divides inversely proportional to capacitance.

Question 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 $100\mathrm{V}$ supply.

Accepted Answer

(a) For capacitors in parallel, total capacitance is sum of individual capacitances:

C_total = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF

(b) Voltage across each capacitor is same: V = 100 V

Charge on each capacitor:
Q₁ = C₁ * V = 2 × 10⁻¹² * 100 = 2 × 10⁻¹⁰ C
Q₂ = 3 × 10⁻¹² * 100 = 3 × 10⁻¹⁰ C
Q₃ = 4 × 10⁻¹² * 100 = 4 × 10⁻¹⁰ C

Therefore, charges are 2 × 10⁻¹⁰ C, 3 × 10⁻¹⁰ C and 4 × 10⁻¹⁰ C respectively. — In parallel, capacitances add. Voltage is same across each capacitor. Charge on each is product of capacitance and voltage.

Question 8

In a parallel plate capacitor with air between the plates, each plate has an area of $6 	imes 10^{-3} \mathrm{~m}^2$ and the distance between the plates is $3 \mathrm{~mm}$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100 \mathrm{~V}$ supply, what is the charge on each plate of the capacitor?

Accepted Answer

Given:
Area, A = 6 × 10⁻³ m²
Distance, d = 3 mm = 3 × 10⁻³ m
Voltage, V = 100 V

Capacitance of parallel plate capacitor:
C = ε₀ * A / d
where ε₀ = 8.854 × 10⁻¹² F/m

Calculate C:
C = (8.854 × 10⁻¹²) * (6 × 10⁻³) / (3 × 10⁻³)
= (8.854 × 10⁻¹²) * 2
= 1.7708 × 10⁻¹¹ F = 17.7 pF

Charge on each plate:
Q = C * V = 1.7708 × 10⁻¹¹ * 100 = 1.7708 × 10⁻⁹ C

Therefore,
Capacitance = 17.7 pF
Charge on each plate = 1.77 nC — Use formula for capacitance of parallel plate capacitor. Charge is product of capacitance and voltage.

Question 9

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.

Accepted Answer

Given:
Thickness of mica sheet, d = 3 mm = 3 × 10⁻³ m
Dielectric constant, K = 6

(a) When the voltage supply remains connected:

The voltage V across the capacitor remains constant.

The capacitance increases by a factor of K when the dielectric is inserted.

If C₀ is the original capacitance, then new capacitance C = K × C₀

Since Q = CV, charge increases by factor K.

Energy stored, U = ½ CV²

Since C increases, energy stored increases by factor K.

(b) When the supply is disconnected:

Charge Q on capacitor remains constant.

Capacitance increases by factor K.

Voltage V = Q/C decreases by factor K.

Energy stored, U = Q²/(2C)

Since C increases, energy stored decreases by factor K.

Hence, energy stored decreases when dielectric is inserted after disconnecting supply. — Step 1: Understand that inserting dielectric increases capacitance by factor K.
Step 2: When voltage is constant (supply connected), charge increases, energy increases.
Step 3: When charge is constant (supply disconnected), voltage decreases, energy decreases.
Step 4: Use formulas for energy stored in capacitor to compare before and after insertion.

Question 10

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Accepted Answer

Given:
Capacitance, C = 12 pF = 12 × 10⁻¹² F
Voltage, V = 50 V

Energy stored, U = ½ CV²
= ½ × 12 × 10⁻¹² × (50)²
= 0.5 × 12 × 10⁻¹² × 2500
= 15 × 10⁻⁹ J
= 15 nJ

Therefore, the electrostatic energy stored in the capacitor is 15 nanojoules. — Step 1: Convert capacitance to farads.
Step 2: Use formula U = ½ CV².
Step 3: Substitute values and calculate energy.

Question 11

A 600pF capacitor is charged by a 200V 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?

Accepted Answer

Given:
Capacitance of each capacitor, C = 600 pF = 600 × 10⁻¹² F
Initial voltage, V = 200 V

Step 1: Calculate initial energy stored in charged capacitor:
U₁ = ½ CV² = 0.5 × 600 × 10⁻¹² × (200)²
= 0.5 × 600 × 10⁻¹² × 40000
= 12 × 10⁻⁶ J = 12 μJ

Step 2: After disconnecting supply, connect to uncharged capacitor of same capacitance.
Total capacitance, C_total = 600 pF + 600 pF = 1200 pF

Charge initially on charged capacitor:
Q = CV = 600 × 10⁻¹² × 200 = 1.2 × 10⁻⁷ C

After connection, charge is shared between two capacitors:
Voltage across both capacitors:
V' = Q / C_total = 1.2 × 10⁻⁷ / (1200 × 10⁻¹²) = 100 V

Step 3: Calculate final energy stored:
U₂ = ½ C_total V'² = 0.5 × 1200 × 10⁻¹² × (100)²
= 0.5 × 1200 × 10⁻¹² × 10000
= 6 × 10⁻⁶ J = 6 μJ

Step 4: Energy lost:
ΔU = U₁ - U₂ = 12 μJ - 6 μJ = 6 μJ

Therefore, 6 microjoules of electrostatic energy is lost in the process. — Step 1: Calculate initial energy using U = ½ CV².
Step 2: Calculate total capacitance after connection.
Step 3: Calculate new voltage after charge sharing.
Step 4: Calculate final energy stored.
Step 5: Find difference to get energy lost.

Question 12

What is electrostatics in physics?

Accepted Answer

Electrostatics is the branch of physics that deals with the study of electric charges at rest. For example, it studies the forces between stationary charges and the electric fields they produce. — Electrostatics focuses on electric charges that are not moving and the interactions between them, including forces, fields, and potential energy. This foundational knowledge is essential for understanding electric phenomena in static conditions.

Question 13

Electrostatic potential is a scalar quantity. What does it represent in an electric field?

Accepted Answer

Electrostatic potential represents the work done in bringing a unit positive charge from infinity to a point in the electric field without acceleration. For example, the potential at a point near a charge indicates the energy per unit charge at that point. — Unlike electric field which is a vector, potential is scalar and simplifies analysis by giving a single value at each point. It measures energy aspects of the field rather than force direction.

Question 14

Define capacitance and state its physical significance.

Accepted Answer

Capacitance is the ability of a system to store electric charge and energy. It is defined as the ratio of the charge stored on the conductor to the potential difference across it. For example, a capacitor with capacitance C stores charge Q when a potential difference V is applied, where C = Q/V. — Capacitance quantifies how much charge a conductor can hold per unit potential difference, which is crucial for designing circuits that store and manage electric energy.

Question 15

Which of the following best describes the relationship between electric field and electrostatic potential?

Accepted Answer

Electric field is the gradient of electrostatic potential and points in the direction of decreasing potential. — Electric field is a vector quantity defined as the negative gradient of the electrostatic potential, meaning it points from higher to lower potential. Potential is scalar, while field has direction and magnitude.

Electrostatic Potential And Capacitance 2.1 Introduction

Electrostatic Potential And Capacitance 2.1 Introduction — Study Notes

2.1 Introduction

Practice Questions — Electrostatic Potential And Capacitance 2.1 Introduction

All 8 Chapters in Physics Part-I