Potential Due to a Continuous Charge Distribution MCQ Quiz - Objective Question with Answer for Potential Due to a Continuous Charge Distribution - Download Free PDF

Concept:

When two charged metallic spheres are brought into contact, they share their charges until both reach the same potential. The potential on the surface of a conducting sphere is directly proportional to the charge and inversely proportional to its radius.

Explanation:

Let the initial charges on the two spheres be Q1 and Q2, and let their respective radii be R1 and R2. When the spheres are connected, they redistribute their charges to reach the same surface potential.

The potential V on a sphere is given by: V = Q / R

Since both spheres attain the same potential after contact:

Q1 / R1 = Q2 / R2

Rearranging this equation:

Q1 / Q2 = R1 / R2

Therefore, the ratio of the final charges on the two spheres after separation is:

Q1 : Q2 = R1 : R2

Hence, the correct option is (4).

CONCEPT:

• Surface charge density (σ) is the quantity of charge (Q) per unit area (A).

σ = Q/A 

so, Charge on the surface = Area of surface (A) × surface charge density (σ)

Electric Potential at the center of a sphere of radius R and charge Q on its surface is given by:

\(V=\frac{1}{4\pi ϵ_0}\frac{Q}{R} \)

where V is the potential at the center, Q is the charge on it, R is the radius of the sphere and ϵ₀ is the electrical permittivity of vacuum.

CALCULATION:

Charge on outer sphere Q₁ = 4πR² σ 

Charge on inner sphere Q₂ = 4πr2 σ 

\(V=\frac{1}{4\pi ϵ_0}\frac{Q_1}{R} +\frac{1}{4\pi ϵ_0}\frac{Q_2}{r}\)

\(V=\frac{1}{4\pi ϵ_0}\frac{4\pi R^2 σ}{R} +\frac{1}{4\pi ϵ_0}\frac{4\pi r^2 σ}{r}\)

\(V=\frac{σ}{ϵ_0}(R+r)\)

so the correct answer is option 1.

CONCEPT:

• Surface charge density (σ) is the quantity of charge (Q) per unit area (A).

σ = Q/A 

so, Charge on the surface = Area of surface (A) × surface charge density (σ)

Electric Potential at the center of a sphere of radius R and charge Q on its surface is given by:

\(V=\frac{1}{4\pi ϵ_0}\frac{Q}{R} \)

where V is the potential at the center, Q is the charge on it, R is the radius of the sphere and ϵ₀ is the electrical permittivity of vacuum.

CALCULATION:

Charge on outer sphere Q₁ = 4πR² σ 

Charge on inner sphere Q₂ = 4πr2 σ 

\(V=\frac{1}{4\pi ϵ_0}\frac{Q_1}{R} +\frac{1}{4\pi ϵ_0}\frac{Q_2}{r}\)

\(V=\frac{1}{4\pi ϵ_0}\frac{4\pi R^2 σ}{R} +\frac{1}{4\pi ϵ_0}\frac{4\pi r^2 σ}{r}\)

\(V=\frac{σ}{ϵ_0}(R+r)\)

so the correct answer is option 1.

CONCEPT:

• Surface charge density (σ) is the quantity of charge (Q) per unit area (A).

σ = Q/A 

so, Charge on the surface = Area of surface (A) × surface charge density (σ)

Electric Potential at the center of a sphere of radius R and charge Q on its surface is given by:

\(V=\frac{1}{4\pi ϵ_0}\frac{Q}{R} \)

where V is the potential at the center, Q is the charge on it, R is the radius of the sphere and ϵ₀ is the electrical permittivity of vacuum.

CALCULATION:

Charge on outer sphere Q₁ = 4πR² σ 

Charge on inner sphere Q₂ = 4πr2 σ 

\(V=\frac{1}{4\pi ϵ_0}\frac{Q_1}{R} +\frac{1}{4\pi ϵ_0}\frac{Q_2}{r}\)

\(V=\frac{1}{4\pi ϵ_0}\frac{4\pi R^2 σ}{R} +\frac{1}{4\pi ϵ_0}\frac{4\pi r^2 σ}{r}\)

\(V=\frac{σ}{ϵ_0}(R+r)\)

so the correct answer is option 1.

The correct answer is option 2) i.e. 2: 1

CONCEPT:

• Electric potential: The electric potential is the difference in potential energy per unit charge between two points in an electric field. 

The electric potential V at a distance r from a point charge Q is given as:

\(V = \frac{kQ}{r}\)

Where k is Coulomb constant.

• The surface charge distribution is when the charge varies along the surface of the conductor. It is given by

\(σ = \frac{dq}{ds}\)

Where dq is the charge and ds is the surface area of the conductor.

CALCULATION:

Let the charges on P and Q be q_P and q_Q and the surface charge densities be σ_P and σ_Q.

\(\sigma_P = \frac{q_P}{4\pi R_P^2}\) and \(\sigma_Q = \frac{q_Q}{4\pi R_Q^2}\)

Since they have the same surface charge densities, σP = σQ

\( \frac{q_P}{4\pi R_P^2} = \frac{q_Q}{4\pi R_Q^2}\)

\(\Rightarrow \frac{q_P}{q_Q} = (\frac{R_P}{R_Q})^2\)

Electric potential, \(V \propto \frac{q}{R}\)

Ratio \(\frac{V_P}{V_Q} = \frac{q_P /R_P}{q_Q/R_Q} =(\frac{R_P}{R_Q})^2 \times \frac{R_Q}{R_P} =\frac{R_P}{R_Q}\)

Given that, RP = 2RQ​

Therefore \(\frac{V_P}{V_Q} = \frac{2R_Q}{R_Q} = \frac{2}{1}\)

CONCEPT:

• Surface charge density (σ) is the quantity of charge (Q) per unit area (A).

σ = Q/A 

so, Charge on the surface = Area of surface (A) × surface charge density (σ)

Electric Potential at the center of a sphere of radius R and charge Q on its surface is given by:

\(V=\frac{1}{4\pi ϵ_0}\frac{Q}{R} \)

where V is the potential at the center, Q is the charge on it, R is the radius of the sphere and ϵ₀ is the electrical permittivity of vacuum.

CALCULATION:

Charge on outer sphere Q₁ = 4πR² σ 

Charge on inner sphere Q₂ = 4πr2 σ 

\(V=\frac{1}{4\pi ϵ_0}\frac{Q_1}{R} +\frac{1}{4\pi ϵ_0}\frac{Q_2}{r}\)

\(V=\frac{1}{4\pi ϵ_0}\frac{4\pi R^2 σ}{R} +\frac{1}{4\pi ϵ_0}\frac{4\pi r^2 σ}{r}\)

\(V=\frac{σ}{ϵ_0}(R+r)\)

so the correct answer is option 1.

Concept:

When two charged metallic spheres are brought into contact, they share their charges until both reach the same potential. The potential on the surface of a conducting sphere is directly proportional to the charge and inversely proportional to its radius.

Explanation:

Let the initial charges on the two spheres be Q1 and Q2, and let their respective radii be R1 and R2. When the spheres are connected, they redistribute their charges to reach the same surface potential.

The potential V on a sphere is given by: V = Q / R

Since both spheres attain the same potential after contact:

Q1 / R1 = Q2 / R2

Rearranging this equation:

Q1 / Q2 = R1 / R2

Therefore, the ratio of the final charges on the two spheres after separation is:

Q1 : Q2 = R1 : R2

Hence, the correct option is (4).

CONCEPT:

• Surface charge density (σ) is the quantity of charge (Q) per unit area (A).

σ = Q/A 

so, Charge on the surface = Area of surface (A) × surface charge density (σ)

Electric Potential at the center of a sphere of radius R and charge Q on its surface is given by:

\(V=\frac{1}{4\pi ϵ_0}\frac{Q}{R} \)

where V is the potential at the center, Q is the charge on it, R is the radius of the sphere and ϵ₀ is the electrical permittivity of vacuum.

CALCULATION:

Charge on outer sphere Q₁ = 4πR² σ 

Charge on inner sphere Q₂ = 4πr2 σ 

\(V=\frac{1}{4\pi ϵ_0}\frac{Q_1}{R} +\frac{1}{4\pi ϵ_0}\frac{Q_2}{r}\)

\(V=\frac{1}{4\pi ϵ_0}\frac{4\pi R^2 σ}{R} +\frac{1}{4\pi ϵ_0}\frac{4\pi r^2 σ}{r}\)

\(V=\frac{σ}{ϵ_0}(R+r)\)

so the correct answer is option 1.

The correct answer is option 2) i.e. 2: 1

CONCEPT:

• Electric potential: The electric potential is the difference in potential energy per unit charge between two points in an electric field. 

The electric potential V at a distance r from a point charge Q is given as:

\(V = \frac{kQ}{r}\)

Where k is Coulomb constant.

• The surface charge distribution is when the charge varies along the surface of the conductor. It is given by

\(σ = \frac{dq}{ds}\)

Where dq is the charge and ds is the surface area of the conductor.

CALCULATION:

Let the charges on P and Q be q_P and q_Q and the surface charge densities be σ_P and σ_Q.

\(\sigma_P = \frac{q_P}{4\pi R_P^2}\) and \(\sigma_Q = \frac{q_Q}{4\pi R_Q^2}\)

Since they have the same surface charge densities, σP = σQ

\( \frac{q_P}{4\pi R_P^2} = \frac{q_Q}{4\pi R_Q^2}\)

\(\Rightarrow \frac{q_P}{q_Q} = (\frac{R_P}{R_Q})^2\)

Electric potential, \(V \propto \frac{q}{R}\)

Ratio \(\frac{V_P}{V_Q} = \frac{q_P /R_P}{q_Q/R_Q} =(\frac{R_P}{R_Q})^2 \times \frac{R_Q}{R_P} =\frac{R_P}{R_Q}\)

Given that, RP = 2RQ​

Therefore \(\frac{V_P}{V_Q} = \frac{2R_Q}{R_Q} = \frac{2}{1}\)