Electrostatic Potential
Electric Potential: The amount of work needed to bring a unit positive charge from infinity to a specific point in an electric field.
Potential Difference: The difference in electric potential between two points, often measured in volts (V).
Equipotential Surfaces
Imagine a surface where the electric potential is the same at every point. This is an equipotential surface. It’s like a contour map, but instead of showing elevation, it shows the electric potential.
Electric Dipole
It’s like a tiny magnet, creating an electric field around it. Dipoles are fundamental to understanding many electrical phenomena, from molecular interactions to the behavior of materials in electric fields.
Capacitance
Capacitor: A device used to store electrical energy in an electric field.
Capacitance: The ability of a capacitor to store charge, measured in farads (F).
Factors Affecting Capacitance:
Plate Area: Larger plates increase capacitance.
Plate Separation: A smaller distance between plates increases capacitance.
Dielectric Material: A dielectric material placed between the plates can increase capacitance.
Capacitors in Circuits
Capacitors in Series: The total capacitance decreases.
Capacitors in Parallel: The total capacitance increases.
Energy Stored in a Capacitor
given,
energy stored in a capacitor
U = (1/2)CV²
where
U denote energy stored,
C denote capacitance,
V denote voltage
across the capacitor.
Capacitors in Series and Parallel:
Series: 1/C_eq = 1/C_1 + 1/C_2 + …
Parallel: C_eq = C_1 + C_2 + …
Energy Stored in a Capacitor: U = (1/2)CV²
Dielectrics
Dielectric Constant: A measure of a material’s ability to reduce the electric field between capacitor plates.
Polarization: The alignment of dipole moments in a dielectric material under an electric field.
This chapter delves into the concepts of electric potential, capacitance, and the role of dielectrics in electric fields.
1 Two charges 5 × 10-8 C and –3 × 10-8C 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.
Ans : Two charges, q₁ = +5.0 x 10⁻¹⁰ C and q₂ = -3.0 x 10⁻¹⁰ C, are separated by a distance of 16 cm.
Part (a):
Find the electric potential at point P, which is located 10 cm from q₁ and 6 cm from q₂.
Solution:
Given,
Potential of q₁ at P
V₁ = kq₁
——-
r₁
=(9 x 10⁹ N m²/C²) x (5 x 10⁻¹⁰ C) / (0.1 m)
= 45 V
The electric potential at point P due to q₂ is given by:
V₂ = kq₂ / r₂ = (9 x 10⁹ N m²/C²) x (-3 x 10⁻¹⁰ C) / (0.06 m) = -45 V
hence, the total electric potential at point P is:
V = V₁ + V₂ = 45 V + (-45 V) = 0 V
So, the electric potential at point P is zero.
Part (b):
Find the distance x from q₁ where the electric potential is zero.
Solution:
Let x be the distance from q₁ where the electric potential is zero.
hen, the distance from q₂ to this point is (16 – x) cm.
Potential due to q₁:
V₁ = kq₁ / x
Potential due to q₁:
V₂ = kq₂ / (16 – x)
Given that the total potential is 0.
V₁ + V₂ = 0
Therefore,
kq₁ / x + kq₂ / (16 – x) = 0
Simplifying and solving for x, we get:
x = 10 cm
So, the electric potential is zero at a distance of 10 cm from q₁.
Two charges, q₁ = +5.0 x 10⁻¹⁰ C and q₂ = -3.0 x 10⁻¹⁰ C, are separated by a distance of 16 cm.
Part (c):
Find the distance from the positive charge where the electric potential is zero.
Solution:
Let’s consider a point P outside the system of charges, at a distance s from the negative charge q₂, where the potential is zero.
Potential of q₁ at P
V₁ = kq₁ / s
Potential of q₂ at P
V₂ = kq₂ / (s – d)
As the total potential is zero
V₁ + V₂ = 0
Therefore,
kq₁ / s + kq₂ / (s – d) = 0
Simplifying and solving for s, we get:
s = 40 cm
So, the electric potential is zero at a distance of 40 cm from the positive charge q₁.
2. A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
Ans :Let’s denote the charge at each vertex as Q = 5 mC = 5 × 10⁻³ C.
The distance from each vertex to the center of the hexagon is the radius of the circumscribing circle. For a regular hexagon, this radius is equal to the side length.
So, the distance between each vertex and the center is 10 cm = 0.1 m.
The electric potential at the center due to one charge Q is given by:
V = kQ/r
Where:
* k =electrostatic constant
(approximately 9 × 10⁹ Nm²/C²)
* Q = charge (5 × 10⁻³ C)
* r = distance from the charge to the center (0.1 m)
Since there are 6 charges, each contributing equally to the potential at the center, the total potential is:
V_total = 6 * (kQ/r) = 6 * (9 × 10⁹ Nm²/C² * 5 × 10⁻³ C / 0.1 m)
= 2.7 × 10⁸ V
hence, the potential at the center of the hexagon approximately 2.7 × 10⁸ volts.
3. Two charges 2 mC and –2 mC 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?
Ans :(a) Equipotential Surface:
An equipotential surface is a surface where the electric potential remains constant throughout.. For a system of two equal and opposite charges, The equipotential surface is a plane bisecting the line joining the charges.
In this case, the equipotential surface is a plane perpendicular to the line AB and passing through the midpoint of AB.
(b)Direction of the Electric Field:
Field lines flow from high to low potential.. On an equipotential surface, the potential is constant. Therefore, the electric field lines must be perpendicular to the equipotential surface at every point.
So, the direction of the electric field at every point on this surface is perpendicular to the plane and parallel to the line joining the two charges.
4. A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C 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?
Ans :We have a spherical conductor with a uniform charge distribution.
Determine electric field at different points: within the sphere, immediately outside its surface, and at a point further away from the sphere
Solution:
a) Inside the Sphere:
For a conductor, all the excess charge resides on its surface. Therefore, there is no charge inside the conductor. Gauss’s Law dictates that the electric field intensity inside a conductor is zero
b) Just Outside the Sphere:
At a point just outside the sphere, we can treat the sphere as a point charge located at its center. The strength of the electric field at a distance r from a point charge Q is:
E = kQ/r²
Where:
k= electrostatic constant (approximately 9 × 10⁹ Nm²/C²)
Q = charge on the sphere (1.6 × 10⁻⁷ C)
r = distance from the center (12 cm = 0.12 m)
Substituting the values:
E = (9 × 10⁹ Nm²/C²) × (1.6 × 10⁻⁷ C)
———————————
(0.12 m)²
≈ 1 × 10⁵ N/C
c) At an 18 cm radius:
Again, we can treat the sphere as a point charge. The distance r has increased to 18 cm, or 0.18 meters.”
E = (9 × 10⁹ Nm²/C²) × (1.6 × 10⁻⁷ C) / (0.18 m)² ≈ 4.4 × 10⁴ N/C
Therefore:
Inside the sphere, the electric field is 0 N/C.
Just outside the sphere, the electric field is approximately 1 × 10⁵ N/C.
At a point 18 cm from the center, the electric field is approximately 4.4 × 10⁴ N/C.
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?
Ans :1. Initial Capacitance:
The initial capacitance (C₁) is given by:
C₁ = ε₀A/d
Where:
ε₀ is the permittivity of free space.
A is the plate area.
d is the initial plate separation.
2. New Capacitance:
When the plate separation is halved (d/2) and a dielectric with a dielectric constant of k is introduced, the new capacitance (C₂) becomes:
C₂ = kε₀A / (d/2) = 2kε₀A/d
3. Ratio of New to Old Capacitance:
Comparing C₂ to C₁, we get:
C₂/C₁ = 2k
4. Calculating New Capacitance:
Given C₁ = 8 pF and k = 6:
C₂ = 2kC₁ = 2 × 6 × 8 pF = 96 pF
Therefore, the new capacitance is 96 pF.
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?
Ans :Understanding the Problem:**
We have three capacitors connected in series. We need to find the equivalent capacitance and the potential difference across each capacitor when connected to a 120 V supply.
Solution:
a) Total Capacitance:
For capacitors in series
1/C_eq = 1/C₁
+ 1/C₂ + 1/C₃
Given that C₁ = C₂ = C₃ = 9 pF, we can substitute:
1/C_eq = 1/9 + 1/9 + 1/9 = 3/9
Therefore, C_eq = 3 pF
b) Potential Difference Across Each Capacitor:
Series capacitors share the same charge.. Let’s denote this charge as Q.
The total voltage across the combination is 120 V. So, we can write:
120 V = Q/C₁ + Q/C₂ + Q/C₃
Given equal capacitors,
120 V = 3Q/C_eq
Substituting C_eq = 3 pF:
120 V = 3Q / (3 × 10⁻¹² F)
Solving for Q:
Q = 120 V × 10⁻¹² C = 1.2 × 10⁻¹⁰ C
Now, we can find the potential difference across each capacitor (V):
V = Q/C = (1.2 × 10⁻¹⁰ C) / (9 × 10⁻¹² F) = 40 V
hence, the potential difference across each capacitor approximately 40 V.
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.
Ans :Understanding the Problem:**
We have three capacitors connected in parallel. We need to find the total capacitance of the combination and the charge on each capacitor when connected to a 100 V supply.
a) Total Capacitance:
The total capacitance of capacitors connected in parallel is the sum of their individual capacitances.
C_total = C₁ + C₂ + C₃
Given:
C₁ = 2 pF
C₂ = 3 pF
C₃ = 4 pF
Substituting the values:
C_total = 2 pF + 3 pF + 4 pF
= 9 pF
b) Charge on Each Capacitor:
In a parallel capacitor arrangement, all capacitors have identical voltage drops across their plates. In this case, the potential difference across each capacitor is 100 V.
given,
charge on a capacitor
Q = CV
For each capacitor:
Capacitor 1:** Q₁
= C₁V
= (2 × 10⁻¹² F)(100 V)
= 2 × 10⁻¹⁰ C
Capacitor 2:** Q₂
= C₂V
= (3 × 10⁻¹² F)(100 V)
= 3 × 10⁻¹⁰ C
Capacitor 3:** Q₃
= C₃V
= (4 × 10⁻¹² F)(100 V)
= 4 × 10⁻¹⁰ C
Therefore, the charges on the capacitors are 2 × 10⁻¹⁰ C, 3 × 10⁻¹⁰ C, and 4 × 10⁻¹⁰ C, respectively.
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
Ans :1. Capacitance Calculation:
The capacitance (C) of a parallel plate capacitor with air as a dielectric is given by:
C = ε₀ * (A/d)
Where:
ε₀ denote permittivity of free space
(8.85 × 10⁻¹² F/m)
A denote area of each plate
(6 × 10⁻³ m²)
d denote distance between the plates
(3 mm = 3 × 10⁻³ m)
Substituting the values:
C = (8.85 × 10⁻¹²) * (6 × 10⁻³) / (3 × 10⁻³) = 1.77 × 10⁻¹¹ F
2. Charge Calculation:
The charge (Q) on a capacitor is related to its capacitance (C) and the voltage across it (V) by the formula:
Q = CV
Given:
C = 1.77 × 10⁻¹¹ F
V = 100 V
Substituting the values:
Q = (1.77 × 10⁻¹¹ F) * 100 V
= 1.77 × 10⁻⁹ C
Therefore, the capacitance of the capacitor is 1.77 × 10⁻¹¹ F, and the charge on each plate is 1.77 × 10⁻⁹ C.
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.
Ans :A parallel-plate capacitor with air as a dielectric has an initial capacitance of 1.771 x 10⁻¹¹ F. A mica sheet (k=6) is inserted.
Part (a):
If a 100 V supply voltage is maintained:
New Capacitance: The introduction of the mica sheet increases the capacitance: C’ = kC = 6 * 1.771 x 10⁻¹¹ F = 1.06 x 10⁻¹⁰ F
New Charge: With a constant voltage and increased capacitance, the charge on the plates increases: Q’ = C’V = 1.06 x 10⁻¹⁰ F * 100 V
= 1.06 x 10⁻⁸ C
Potential Difference: The potential difference remains constant at 100 V, as the voltage source is still connected.
Part (b):
If the supply voltage is removed after the mica sheet is inserted:
Charge: The charge stored on the plates remains the same
Q = 1.771 x 10⁻⁹ C
New Potential Difference: With the same charge but increased capacitance, the potential difference decreases: V’ = Q / C’ = (1.771 x 10⁻⁹ C) / (1.06 x 10⁻¹⁰ F)
= 16.7 V
10. A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Ans :Given,
formula for the energy stored in a capacitor
U = (1/2)CV²
Where:
* U denote stored energy
* C denote capacitance (12 pF = 12 × 10⁻¹² F)
* V = voltage across the capacitor
(50 V)
Substituting the values:
U =
(1/2) × (12 × 10⁻¹²) F × (50 V)²
Calculating the energy:
U = 1.5 × 10⁻⁸ J
hence, the electrostatic energy stored in the capacitor is 1.5 × 10⁻⁸ Joules.
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?
Ans : We have two capacitors:
1. Initially charged to 200V
2. Initially uncharged
When they’re connected, charge will redistribute until both capacitors have the same potential difference.
Initial Energy:
(U₁) denote initial energy which stored in the 1st capacitor=
U₁
= (1/2)CV²
= (1/2)(600 × 10⁻¹²) × (200)²
= 1.2 × 10⁻⁵ J
Final Energy:
Parallel capacitors add up is
C_total
= C₁ + C₂
= 600 pF + 600 pF
= 1200 pF
The total charge remains constant:
Q_total
= C₁V₁
= (600 × 10⁻¹²) × 200
= 1.2 × 10⁻⁷ C
The final voltage across each capacitor is:
V_final
= Q_total / C_total
= (1.2 × 10⁻⁷ C) / (1200 × 10⁻¹²) F
= 100 V
The final energy (U₂) stored in both capacitors is:
U₂
= (1/2)C_totalV_final²
= (1/2)(1200 × 10⁻¹²) × (100)²
= 6 × 10⁻⁷ J
Energy Loss:
The energy lost in the process is the difference between the initial and final energies:
Energy loss = U₁ – U₂ = 1.2 × 10⁻⁵ J – 6 × 10⁻⁷ J
= 1.14 × 10⁻⁵ J
hence, 1.14 × 10⁻⁵ Joules** of electrostatic energy is lost in the process.
