The Electric Field Due to a Charged Disk MCQ Quiz - Objective Question with Answer for The Electric Field Due to a Charged Disk - Download Free PDF
Last updated on Mar 28, 2025
Latest The Electric Field Due to a Charged Disk MCQ Objective Questions
The Electric Field Due to a Charged Disk Question 1:
An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density σ0. Choose the correct option from the following.
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 1 Detailed Solution
Calculation:
Here
∴ The potential energy of the dipole is at a minimum, and torque is zero.
The Electric Field Due to a Charged Disk Question 2:
A small electric dipole
If released from rest, then which of the following statement(s) is (are) correct?
[ε0 is the permittivity of free space.]
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 2 Detailed Solution
Calculation:
∴
For r = 2R
Also, for r = 10R
∴ Options 2 & 4 are correct.
The Electric Field Due to a Charged Disk Question 3:
Electric field at a distance r from an infinite thin plane sheet of uniform surface charge density σ is
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 3 Detailed Solution
Explanation:
The electric field at a distance r from an infinite thin plane sheet of uniform surface charge density
where
Thus, option '2' is correct.
The Electric Field Due to a Charged Disk Question 4:
An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density +σ. The electron at t = 0 is at a distance of 1 m from S and has a speed of 1 m/s. The maximum value of σ if the electron strikes S at t = l s is α
Answer (Detailed Solution Below) 8
The Electric Field Due to a Charged Disk Question 4 Detailed Solution
Calculation:
Given that,
u = 1 m/s;
t = 1 s
S = –1 m
Using
∴ α = 8
The Electric Field Due to a Charged Disk Question 5:
The electric potential at the centre of two concentric half rings of radii R1 and R2, having same linear charge density λ is :
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 5 Detailed Solution
Calculation:
Electric potential due to circular arc (linear charge)
VC = V1 + V2
⇒
⇒
∴ The correct answer is Option (1):
Top The Electric Field Due to a Charged Disk MCQ Objective Questions
Calculate the electric field intensity at a distance of 10 cm from a large metallic sheet of area 400 m2. The charge of 26.55 × 10-4 C is distributed over the large metallic sheet.
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 6 Detailed Solution
Download Solution PDFCONCEPT:
- Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e.
- But we know that Electrical flux through a closed surface is
Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.:
- The electric field at a point due to an infinite sheet of charge is
Where ϵo = Absolute electrical permittivity of free space, E = Electric field, and σ = surface charge density.
CALCULATION:
Given - Charge (q) = 26.55 × 10-4 C, A = 400 m2 and r = 10 cm = 10-1 m
- The electric field at a point due to an infinite sheet of charge is
Electric field intensity due to thin infinite parallel sheets of charge in region 1 is
( a field pointing from left to right is taken as positive and the one pointing from right to left is taken as negative. )
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 7 Detailed Solution
Download Solution PDFCONCEPT:
- Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e.
- But we know that Electrical flux through a closed surface is
Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.
- The electric field at a point due to an infinite sheet of charge is
Where ϵo = Absolute electrical permittivity of free space, E = Electric field, and σ = surface charge density.
EXPLANATION:
Let σ1 = Uniform surface density of charge on A, σ2 = Uniform surface density of charge on B, E1, E2 = Electric field intensities at a point due to charged sheet A and B respectively.
Here,
- The arrangement shows three regions I, II, and III.
- We apply the superposition principle to calculate the net field intensity in the three regions. As a matter of convention, a field pointing from left to right is taken as positive and the one pointing from right to left is taken as negative.
We assume, σ1 > σ2 > 0
In region 1:
In region 2:
In region 3:
Important Points
- The electric field between them is given by:
The electric field due to a thin spherical shell having a charge 'q', is given as _______________, where 'r' is the distance of the point from the center of the shell, (outside the shell). ('εo' is the permittivity of free space)
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 8 Detailed Solution
Download Solution PDFCONCEPT:
- Gauss Law: According to gauss’s law, total electric flux through a closed surface enclosing a charge is 1/ϵ0 times the magnitude of charge enclosed.
Where, Φ = electric flux, Qin = charge enclosed the sphere, ϵ0 = permittivity of space (8.85 × 10-12 C2/Nm2), dS = surface area
The electric field due to a thin spherical shell having a charge q:
Enclosed charge (Qin) = q
Area (S) = 4 π r2
Use Gauss' law:
⇒ E. (S) = q/ϵ0
⇒ E.4 π r2 = q/ϵ0
Where K is a constant = 1/(4πϵ0) = 9× 109 Nm2/C2, q is charge and r is the distance from charge particle.
EXPLANATION:
The electric field due to a thin spherical shell having a charge 'q', is given as:
E = q/(4πεor2)
So option 3 is correct
The electric field at distance r from a uniformly charged infinite sheet of charge density σ will be :
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 9 Detailed Solution
Download Solution PDFCONCEPT:
- Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e.
- But we know that Electrical flux through a closed surface is
Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.
EXPLANATION:
- The electric field at a point due to infinite sheet of charge is
Where ϵo = Absolute electrical permittivity of free space, E = Electric field, and σ = surface charge density.
- Thus, the field is uniform and does not depend on the distance from the plane sheet of charge. Therefore option 1 is correct.
The Electric Field Due to a Charged Disk Question 10:
Calculate the electric field intensity at a distance of 10 cm from a large metallic sheet of area 400 m2. The charge of 26.55 × 10-4 C is distributed over the large metallic sheet.
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 10 Detailed Solution
CONCEPT:
- Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e.
- But we know that Electrical flux through a closed surface is
Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.:
- The electric field at a point due to an infinite sheet of charge is
Where ϵo = Absolute electrical permittivity of free space, E = Electric field, and σ = surface charge density.
CALCULATION:
Given - Charge (q) = 26.55 × 10-4 C, A = 400 m2 and r = 10 cm = 10-1 m
- The electric field at a point due to an infinite sheet of charge is
The Electric Field Due to a Charged Disk Question 11:
Electric field intensity due to thin infinite parallel sheets of charge in region 1 is
( a field pointing from left to right is taken as positive and the one pointing from right to left is taken as negative. )
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 11 Detailed Solution
CONCEPT:
- Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e.
- But we know that Electrical flux through a closed surface is
Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.
- The electric field at a point due to an infinite sheet of charge is
Where ϵo = Absolute electrical permittivity of free space, E = Electric field, and σ = surface charge density.
EXPLANATION:
Let σ1 = Uniform surface density of charge on A, σ2 = Uniform surface density of charge on B, E1, E2 = Electric field intensities at a point due to charged sheet A and B respectively.
Here,
- The arrangement shows three regions I, II, and III.
- We apply the superposition principle to calculate the net field intensity in the three regions. As a matter of convention, a field pointing from left to right is taken as positive and the one pointing from right to left is taken as negative.
We assume, σ1 > σ2 > 0
In region 1:
In region 2:
In region 3:
Important Points
- The electric field between them is given by:
The Electric Field Due to a Charged Disk Question 12:
The electric field due to a thin spherical shell having a charge 'q', is given as _______________, where 'r' is the distance of the point from the center of the shell, (outside the shell). ('εo' is the permittivity of free space)
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 12 Detailed Solution
CONCEPT:
- Gauss Law: According to gauss’s law, total electric flux through a closed surface enclosing a charge is 1/ϵ0 times the magnitude of charge enclosed.
Where, Φ = electric flux, Qin = charge enclosed the sphere, ϵ0 = permittivity of space (8.85 × 10-12 C2/Nm2), dS = surface area
The electric field due to a thin spherical shell having a charge q:
Enclosed charge (Qin) = q
Area (S) = 4 π r2
Use Gauss' law:
⇒ E. (S) = q/ϵ0
⇒ E.4 π r2 = q/ϵ0
Where K is a constant = 1/(4πϵ0) = 9× 109 Nm2/C2, q is charge and r is the distance from charge particle.
EXPLANATION:
The electric field due to a thin spherical shell having a charge 'q', is given as:
E = q/(4πεor2)
So option 3 is correct
The Electric Field Due to a Charged Disk Question 13:
The electric field at distance r from a uniformly charged infinite sheet of charge density σ will be :
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 13 Detailed Solution
CONCEPT:
- Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e.
- But we know that Electrical flux through a closed surface is
Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.
EXPLANATION:
- The electric field at a point due to infinite sheet of charge is
Where ϵo = Absolute electrical permittivity of free space, E = Electric field, and σ = surface charge density.
- Thus, the field is uniform and does not depend on the distance from the plane sheet of charge. Therefore option 1 is correct.
The Electric Field Due to a Charged Disk Question 14:
An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density σ0. Choose the correct option from the following.
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 14 Detailed Solution
Calculation:
Here
∴ The potential energy of the dipole is at a minimum, and torque is zero.
The Electric Field Due to a Charged Disk Question 15:
The electric potential at the centre of two concentric half rings of radii R1 and R2, having same linear charge density λ is :
Answer (Detailed Solution Below)
The Electric Field Due to a Charged Disk Question 15 Detailed Solution
Calculation:
Electric potential due to circular arc (linear charge)
VC = V1 + V2
⇒
⇒
∴ The correct answer is Option (1):