Electrostatic Potential And CapacitanceClass 12 Physics NCERT Solutions
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Q1EXERCISES
2.1 Two charges 5 × 10⁻⁸ C and –3 × 10⁻⁸ 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.
Solution
Let the two charges be q₁ = 5 × 10⁻⁸ C and q₂ = –3 × 10⁻⁸ C. The distance between them is d = 16 cm = 0.16 m.
Let us place the charge q₁ at the origin (x = 0). Then q₂ is located at x = 0.16 m. Let P be the point on the line where the total electric potential is zero. The potential at P is the sum of the potentials due to q₁ and q₂.
Case 1: The point P lies between the two charges.
Let the distance of P from q₁ be x. Then its distance from q₂ will be (0.16 – x).
The potential at P is V = V₁ + V₂ = 0.
V = (1 / 4πε₀) * [q₁/x + q₂/(0.16 - x)] = 0
(5 × 10⁻⁸) / x + (–3 × 10⁻⁸) / (0.16 – x) = 0
5/x = 3 / (0.16 – x)
5(0.16 – x) = 3x
0.8 – 5x = 3x
8x = 0.8
x = 0.1 m = 10 cm
So, the potential is zero at a point 10 cm from the positive charge, between the two charges.
Case 2: The point P lies outside the two charges.
The potential can be zero only on the side of the smaller charge (in magnitude), which is q₂. Let the point P be at a distance x from q₁.
Its distance from q₂ will be (x – 0.16).
The potential at P is V = V₁ + V₂ = 0.
V = (1 / 4πε₀) * [q₁/x + q₂/(x - 0.16)] = 0
(5 × 10⁻⁸) / x + (–3 × 10⁻⁸) / (x – 0.16) = 0
5/x = 3 / (x – 0.16)
5(x – 0.16) = 3x
5x – 0.8 = 3x
2x = 0.8
x = 0.4 m = 40 cm
So, the potential is also zero at a point 40 cm from the positive charge and 24 cm from the negative charge, on the extended line.
Thus, the two points where the electric potential is zero are at 10 cm and 40 cm from the charge 5 × 10⁻⁸ C.
Q2EXERCISES
2.2 A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Solution
Given:
- Charge at each vertex, q = 5 μC = 5 × 10⁻⁶ C
- Side of the regular hexagon, a = 10 cm = 0.1 m
In a regular hexagon, the distance from the centre to each vertex is equal to the side length.
So, the distance of the centre from each charge is r = a = 0.1 m.
The electric potential at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.
The potential at the centre due to one charge is:
V' = (1 / 4πε₀) * (q / r)
Since there are six identical charges placed at the same distance from the centre, the total potential V at the centre will be 6 times the potential due to a single charge.
V = 6 × V' = 6 × (1 / 4πε₀) * (q / r)
We know that 1 / 4πε₀ = 9 × 10⁹ Nm²C⁻².
V = 6 × (9 × 10⁹) × (5 × 10⁻⁶) / 0.1
V = 6 × 9 × 5 × 10⁹⁻⁶⁺¹
V = 270 × 10⁴ V
V = 2.7 × 10⁶ V
Therefore, the potential at the centre of the hexagon is 2.7 × 10⁶ V.
Q3EXERCISES
2.3 Two charges 2 μC and –2 μC 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?
Solution
The system consists of two equal and opposite charges separated by a distance, which constitutes an electric dipole.
(a) Equipotential Surface:
An equipotential surface is a surface where the electric potential is the same at every point. For an electric dipole, the potential at any point on the equatorial plane (the plane that is perpendicular to the line joining the two charges and passes through its midpoint) is zero. This is because any point on this plane is equidistant from both charges. The potential due to the positive charge is equal and opposite to the potential due to the negative charge, and their sum is zero.
Thus, the plane perpendicular to the line AB and passing through its midpoint is an equipotential surface with a potential of zero.
(b) Direction of Electric Field:
The electric field is always directed perpendicular to the equipotential surface. The direction of the electric field is from a region of higher potential to a region of lower potential. On this equipotential surface (the equatorial plane), the electric field lines are directed from the positive charge (2 μC) to the negative charge (–2 μC). Therefore, the direction of the electric field at every point on this surface is parallel to the line joining the charges (the dipole axis) and points from A to B.
Q4EXERCISES
2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ 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 of the sphere?
Solution
Given:
- Radius of the spherical conductor, R = 12 cm = 0.12 m
- Charge on the conductor, Q = 1.6 × 10⁻⁷ C
(a) Inside the sphere:
The electrostatic field inside a charged conductor is always zero. This is because the charges reside on the surface and rearrange themselves in such a way that the net field inside is nullified.
Therefore, the electric field inside the sphere is E = 0.
(b) Just outside the sphere:
For a point just outside the surface (r = R), the electric field is given by the formula for a point charge located at the centre.
E = (1 / 4πε₀) * (Q / R²)
E = (9 × 10⁹ Nm²C⁻²) × (1.6 × 10⁻⁷ C) / (0.12 m)²
E = (9 × 1.6 × 10²) / 0.0144
E = (14.4 × 10²) / 0.0144
E = 100 × 10² = 10⁴ N/C
The direction of the electric field is radially outward since the charge is positive.
(c) At a point 18 cm from the centre of the sphere:
The distance of the point from the centre is r = 18 cm = 0.18 m. Since this point is outside the sphere, the electric field is calculated as if the entire charge were concentrated at the centre.
E = (1 / 4πε₀) * (Q / r²)
E = (9 × 10⁹ Nm²C⁻²) × (1.6 × 10⁻⁷ C) / (0.18 m)²
E = (9 × 1.6 × 10²) / 0.0324
E = (14.4 × 10²) / 0.0324
E ≈ 4.44 × 10⁴ N/C
The direction of the electric field is radially outward.
Q5EXERCISES
2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 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?
Solution
Let the initial capacitance of the parallel plate capacitor be C₀.
Given, C₀ = 8 pF = 8 × 10⁻¹² F.
The capacitance of a parallel plate capacitor with air (or vacuum) as the medium is given by:
C₀ = ε₀A / d
where A is the area of the plates and d is the distance between them.
According to the problem, the new conditions are:
- New distance between the plates, d' = d / 2
- The space is filled with a substance of dielectric constant, K = 6
The capacitance of a capacitor with a dielectric medium is given by:
C = Kε₀A / d'
Now, we substitute the new values into this formula:
C = (6) * ε₀A / (d / 2)
C = 12 * (ε₀A / d)
Since C₀ = ε₀A / d, we can write:
C = 12 * C₀
Substituting the value of C₀:
C = 12 × 8 pF
C = 96 pF
Therefore, the new capacitance will be 96 pF.
Q6EXERCISES
2.6 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?
Solution
Given:
- Capacitance of each capacitor, C₁ = C₂ = C₃ = 9 pF = 9 × 10⁻¹² F
- Supply voltage, V = 120 V
(a) Total capacitance of the combination:
When capacitors are connected in series, the equivalent capacitance (C_eq) is given by the formula:
1/C_eq = 1/C₁ + 1/C₂ + 1/C₃
1/C_eq = 1/9 + 1/9 + 1/9
1/C_eq = 3/9 = 1/3
C_eq = 3 pF
The total capacitance of the combination is 3 pF.
(b) Potential difference across each capacitor:
In a series combination, the charge (Q) on each capacitor is the same and is equal to the total charge supplied by the battery.
Q = C_eq × V
Q = (3 × 10⁻¹² F) × (120 V)
Q = 360 × 10⁻¹² C
The potential difference (V') across each capacitor is given by V' = Q / C.
Since all three capacitors have the same capacitance (C = 9 pF), the potential difference across each will be the same.
V₁ = Q / C₁ = (360 × 10⁻¹² C) / (9 × 10⁻¹² F) = 40 V
V₂ = Q / C₂ = (360 × 10⁻¹² C) / (9 × 10⁻¹² F) = 40 V
V₃ = Q / C₃ = (360 × 10⁻¹² C) / (9 × 10⁻¹² F) = 40 V
Thus, the potential difference across each capacitor is 40 V. (Check: 40V + 40V + 40V = 120V).
Q7EXERCISES
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.
Solution
Given:
- Capacitances: C₁ = 2 pF, C₂ = 3 pF, C₃ = 4 pF
- Supply voltage, V = 100 V
(a) Total capacitance of the combination:
When capacitors are connected in parallel, the equivalent capacitance (C_eq) is the sum of the individual capacitances.
C_eq = C₁ + C₂ + C₃
C_eq = 2 pF + 3 pF + 4 pF
C_eq = 9 pF
The total capacitance of the combination is 9 pF.
(b) Charge on each capacitor:
In a parallel combination, the potential difference across each capacitor is the same and is equal to the supply voltage, V = 100 V.
The charge (Q) on a capacitor is given by the formula Q = C × V.
-
Charge on the first capacitor (Q₁): Q₁ = C₁ × V = (2 × 10⁻¹² F) × (100 V) = 200 × 10⁻¹² C = 200 pC
-
Charge on the second capacitor (Q₂): Q₂ = C₂ × V = (3 × 10⁻¹² F) × (100 V) = 300 × 10⁻¹² C = 300 pC
-
Charge on the third capacitor (Q₃): Q₃ = C₃ × V = (4 × 10⁻¹² F) × (100 V) = 400 × 10⁻¹² C = 400 pC
Therefore, the charges on the capacitors are 200 pC, 300 pC, and 400 pC, respectively.
Q8EXERCISES
2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻³ 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 of the capacitor?
Solution
Given:
- Area of each plate, A = 6 × 10⁻³ m²
- Distance between the plates, d = 3 mm = 3 × 10⁻³ m
- Permittivity of free space, ε₀ ≈ 8.85 × 10⁻¹² F/m
- Supply voltage, V = 100 V
1. Calculation of Capacitance:
The capacitance (C) of a parallel plate capacitor with air as the dielectric is given by the formula:
C = ε₀A / d
C = (8.85 × 10⁻¹² F/m) × (6 × 10⁻³ m²) / (3 × 10⁻³ m)
C = (8.85 × 10⁻¹²) × 2 F
C = 17.7 × 10⁻¹² F
C = 17.7 pF
So, the capacitance of the capacitor is 17.7 pF.
2. Calculation of Charge:
The charge (Q) on each plate of the capacitor is given by the formula:
Q = C × V
Q = (17.7 × 10⁻¹² F) × (100 V)
Q = 17.7 × 10⁻¹⁰ C
Q = 1.77 × 10⁻⁹ C or 1.77 nC
Therefore, the charge on each plate of the capacitor is 1.77 × 10⁻⁹ C.
Q9EXERCISES
2.9 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.
Solution
From Exercise 2.8, we have:
- Initial capacitance with air, C₀ = 17.7 pF
- Initial voltage, V₀ = 100 V
- Initial charge, Q₀ = 1.77 × 10⁻⁹ C
A mica sheet of dielectric constant K = 6 is inserted, completely filling the space between the plates.
The new capacitance (C) will be:
C = K × C₀ = 6 × 17.7 pF = 106.2 pF
(a) While the voltage supply remained connected:
When the voltage supply remains connected, the potential difference across the capacitor plates remains constant at V = V₀ = 100 V.
- Capacitance: The capacitance increases to C = 106.2 pF.
- Charge: Since the capacitance has increased while the voltage is constant, the charge on the plates will increase. The new charge Q is: Q = C × V = (106.2 × 10⁻¹² F) × (100 V) = 1.062 × 10⁻⁸ C. The battery supplies more charge to the capacitor.
- Electric Field: The electric field E = V/d remains the same because both V and d are constant.
(b) After the supply was disconnected:
When the supply is disconnected, the capacitor becomes an isolated system. The charge on its plates remains constant at Q = Q₀ = 1.77 × 10⁻⁹ C.
- Capacitance: The capacitance increases to C = 106.2 pF due to the insertion of the mica sheet.
- Potential Difference: Since the charge is constant and capacitance has increased, the potential difference across the plates will decrease. The new voltage V' is: V' = Q / C = (1.77 × 10⁻⁹ C) / (106.2 × 10⁻¹² F) ≈ 16.67 V.
- Electric Field: The electric field E' = V'/d will also decrease. E' = (16.67 V) / (3 × 10⁻³ m) ≈ 5.56 × 10³ V/m. This is because the induced field in the dielectric opposes the original field.
Q10EXERCISES
2.10 A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
Solution
Given:
- Capacitance, C = 12 pF = 12 × 10⁻¹² F
- Voltage, V = 50 V
The electrostatic energy (U) stored in a capacitor is given by the formula:
U = (1/2)CV²
Substituting the given values:
U = (1/2) × (12 × 10⁻¹² F) × (50 V)²
U = (6 × 10⁻¹² F) × (2500 V²)
U = 15000 × 10⁻¹² J
U = 1.5 × 10⁻⁸ J
Therefore, the electrostatic energy stored in the capacitor is 1.5 × 10⁻⁸ J.
Q11EXERCISES
2.11 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?
Solution
Given:
- Capacitance of the first capacitor, C₁ = 600 pF = 600 × 10⁻¹² F
- Initial voltage, V₁ = 200 V
- Capacitance of the second capacitor, C₂ = 600 pF = 600 × 10⁻¹² F
Step 1: Calculate the initial energy stored.
The initial energy (U_initial) is stored only in the first capacitor.
U_initial = (1/2)C₁V₁²
U_initial = (1/2) × (600 × 10⁻¹² F) × (200 V)²
U_initial = (300 × 10⁻¹²) × (40000) J
U_initial = 12 × 10⁶ × 10⁻¹² J = 1.2 × 10⁻⁵ J
Step 2: Calculate the state after connecting the capacitors.
When the charged capacitor is connected to the uncharged capacitor, they are in a parallel combination. The charge will redistribute until they reach a common potential (V_common).
The initial charge on the first capacitor is Q = C₁V₁.
Q = (600 × 10⁻¹² F) × (200 V) = 1.2 × 10⁻⁷ C
This total charge Q is now shared between the two capacitors. The equivalent capacitance of the parallel combination is C_eq = C₁ + C₂.
C_eq = 600 pF + 600 pF = 1200 pF = 1200 × 10⁻¹² F
The common potential is V_common = Q / C_eq.
V_common = (1.2 × 10⁻⁷ C) / (1200 × 10⁻¹² F) = 100 V
Step 3: Calculate the final energy stored.
The final energy (U_final) of the system is:
U_final = (1/2)C_eq(V_common)²
U_final = (1/2) × (1200 × 10⁻¹² F) × (100 V)²
U_final = (600 × 10⁻¹²) × (10000) J
U_final = 6 × 10⁶ × 10⁻¹² J = 0.6 × 10⁻⁵ J
Step 4: Calculate the energy lost.
The energy lost in the process is the difference between the initial and final energy.
Energy Loss = U_initial – U_final
Energy Loss = (1.2 × 10⁻⁵ J) – (0.6 × 10⁻⁵ J)
Energy Loss = 0.6 × 10⁻⁵ J = 6 × 10⁻⁶ J
This energy is dissipated as heat in the connecting wires and as electromagnetic radiation during the transient current flow.