"MCQ Test on Basic Concepts of Electric Charges and Fields"
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1. A soap bubble is given a negative charge, then its radius:
A. Decreases
B. Increases
C. Remains unchanged
D. Nothing can be predicted as information is insufficient
The correct answer is B. Increases. Here's how it works:
Step 1: Effect of Negative Charge
When a soap bubble is given a negative charge, the charges spread uniformly on the surface due to repulsion.
Step 2: Surface Tension and Pressure
This repulsion increases the surface tension and the internal pressure of the bubble, causing the bubble to expand.
Conclusion:
As a result, the radius of the bubble increases.
Step 1: Effect of Negative Charge
When a soap bubble is given a negative charge, the charges spread uniformly on the surface due to repulsion.
Step 2: Surface Tension and Pressure
This repulsion increases the surface tension and the internal pressure of the bubble, causing the bubble to expand.
Conclusion:
As a result, the radius of the bubble increases.
2. A body can be negatively charged by:
A. Giving excess of electrons to it
B. Removing some electrons from it
C. Giving some protons to it
D. Removing some neutrons from it
The correct answer is A. Giving excess of electrons to it. Here's why:
Step 1: Concept of Negative Charge
A body becomes negatively charged when it has more electrons than protons.
Step 2: How to Charge a Body Negatively
By giving excess electrons to the body, the number of negative charges (electrons) increases, resulting in a net negative charge.
Conclusion:
Therefore, a body can be negatively charged by adding extra electrons to it.
Step 1: Concept of Negative Charge
A body becomes negatively charged when it has more electrons than protons.
Step 2: How to Charge a Body Negatively
By giving excess electrons to the body, the number of negative charges (electrons) increases, resulting in a net negative charge.
Conclusion:
Therefore, a body can be negatively charged by adding extra electrons to it.
3. An attractive force between two neutrons is due to:
A. Electrostatic and gravitational
B. Electrostatic and nuclear
C. Gravitational and nuclear
D. Some other forces like Van der Waals
The correct answer is C. Gravitational and nuclear. Here's why:
Step 1: Forces Acting on Neutrons
Neutrons are electrically neutral, so there is no electrostatic force between them. However, the nuclear force, which is much stronger than gravity, binds them together at short ranges.
Step 2: Gravitational Force
Gravitational force exists between any two masses, including neutrons, but is extremely weak compared to the nuclear force.
Conclusion:
Therefore, the attractive force between two neutrons is mainly due to gravitational and nuclear forces.
Step 1: Forces Acting on Neutrons
Neutrons are electrically neutral, so there is no electrostatic force between them. However, the nuclear force, which is much stronger than gravity, binds them together at short ranges.
Step 2: Gravitational Force
Gravitational force exists between any two masses, including neutrons, but is extremely weak compared to the nuclear force.
Conclusion:
Therefore, the attractive force between two neutrons is mainly due to gravitational and nuclear forces.
4. Two particles of equal mass \( m \) and charge \( q \) are placed at a distance of 16 cm. They do not experience any force. The value of \( \frac{q}{m} \) is:
A. \( 1.5 \times 10^{-7} \, \text{C/kg} \)
B. \( 4.2 \times 10^{-8} \, \text{C/kg} \)
C. \( 2.7 \times 10^{-11} \, \text{C/kg} \)
D. \( 3.6 \times 10^{-12} \, \text{C/kg} \)
The correct answer is C. \( 2.7 \times 10^{-11} \, \text{C/kg} \). Here's how to calculate it:
Step 1: Condition for No Force
Since the particles do not experience any net force, the gravitational force between the two particles must be balanced by the electrostatic repulsion between their charges.
Step 2: Gravitational Force
The gravitational force between two masses \( m \) at a distance \( r \) is given by: \[ F_{\text{gravity}} = \frac{G m^2}{r^2} \]
Step 3: Electrostatic Force
The electrostatic force between two charges \( q \) at the same distance \( r \) is: \[ F_{\text{electrostatic}} = \frac{k q^2}{r^2} \]
Step 4: Equating the Forces
For no net force, these two forces must be equal: \[ \frac{G m^2}{r^2} = \frac{k q^2}{r^2} \] The distance \( r \) cancels out, and rearranging the equation gives: \[ \frac{q}{m} = \sqrt{\frac{G}{k}} \]
Step 5: Substituting Constants
Using \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) and \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \): \[ \frac{q}{m} = \sqrt{\frac{6.67 \times 10^{-11}}{9 \times 10^9}} = 2.7 \times 10^{-11} \, \text{C/kg} \]
Conclusion:
The value of \( \frac{q}{m} \) is approximately \( 2.7 \times 10^{-11} \, \text{C/kg} \).
Step 1: Condition for No Force
Since the particles do not experience any net force, the gravitational force between the two particles must be balanced by the electrostatic repulsion between their charges.
Step 2: Gravitational Force
The gravitational force between two masses \( m \) at a distance \( r \) is given by: \[ F_{\text{gravity}} = \frac{G m^2}{r^2} \]
Step 3: Electrostatic Force
The electrostatic force between two charges \( q \) at the same distance \( r \) is: \[ F_{\text{electrostatic}} = \frac{k q^2}{r^2} \]
Step 4: Equating the Forces
For no net force, these two forces must be equal: \[ \frac{G m^2}{r^2} = \frac{k q^2}{r^2} \] The distance \( r \) cancels out, and rearranging the equation gives: \[ \frac{q}{m} = \sqrt{\frac{G}{k}} \]
Step 5: Substituting Constants
Using \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) and \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \): \[ \frac{q}{m} = \sqrt{\frac{6.67 \times 10^{-11}}{9 \times 10^9}} = 2.7 \times 10^{-11} \, \text{C/kg} \]
Conclusion:
The value of \( \frac{q}{m} \) is approximately \( 2.7 \times 10^{-11} \, \text{C/kg} \).
5. When the distance between the charged particles is halved, the force between them becomes:
A. One-fourth
B. Half
C. Double
D. Four times
The correct answer is D. Four times. Here's the reasoning:
Step 1: Coulomb's Law
The electrostatic force between two charged particles is given by Coulomb's law: \[ F = \frac{k q_1 q_2}{r^2} \] where \( F \) is the force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between them, and \( k \) is Coulomb's constant.
Step 2: Effect of Halving the Distance
If the distance \( r \) is halved, the force becomes: \[ F' = \frac{k q_1 q_2}{\left(\frac{r}{2}\right)^2} = \frac{k q_1 q_2}{\frac{r^2}{4}} = 4 \times \frac{k q_1 q_2}{r^2} = 4F \]
Conclusion:
Therefore, when the distance between the particles is halved, the force between them increases by four times.
Step 1: Coulomb's Law
The electrostatic force between two charged particles is given by Coulomb's law: \[ F = \frac{k q_1 q_2}{r^2} \] where \( F \) is the force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between them, and \( k \) is Coulomb's constant.
Step 2: Effect of Halving the Distance
If the distance \( r \) is halved, the force becomes: \[ F' = \frac{k q_1 q_2}{\left(\frac{r}{2}\right)^2} = \frac{k q_1 q_2}{\frac{r^2}{4}} = 4 \times \frac{k q_1 q_2}{r^2} = 4F \]
Conclusion:
Therefore, when the distance between the particles is halved, the force between them increases by four times.
6. A point charge is placed at the center of a spherical conducting shell. The electric field inside the shell (but outside the point charge) is:
A. Zero everywhere
B. Constant and non-zero
C. Radially directed and non-zero
D. Radially directed but zero
The correct answer is C. Radially directed and non-zero. Here's why:
Step 1: Gauss's Law
According to Gauss's Law, the electric field \( E \) at a distance \( r \) from a point charge \( q \) is: \[ E = \frac{q}{4 \pi \epsilon_0 r^2} \]
Step 2: Field in a Spherical Shell
In a spherical conducting shell, the electric field inside the conductor is zero. However, the electric field in the region between the point charge and the inner surface of the shell is non-zero and radially directed due to the point charge.
Conclusion:
The electric field inside the shell but outside the point charge is radially directed and follows the inverse square law.
Step 1: Gauss's Law
According to Gauss's Law, the electric field \( E \) at a distance \( r \) from a point charge \( q \) is: \[ E = \frac{q}{4 \pi \epsilon_0 r^2} \]
Step 2: Field in a Spherical Shell
In a spherical conducting shell, the electric field inside the conductor is zero. However, the electric field in the region between the point charge and the inner surface of the shell is non-zero and radially directed due to the point charge.
Conclusion:
The electric field inside the shell but outside the point charge is radially directed and follows the inverse square law.
7. The number of electrons in one coulomb of charge will be:
A. \( 6.25 \times 10^{18} \) electrons
B. \( 1.6 \times 10^{-19} \) electrons
C. \( 9.1 \times 10^{-31} \) electrons
D. \( 1 \times 10^{6} \) electrons
The correct answer is A. \( 6.25 \times 10^{18} \) electrons. Here's the calculation:
Step 1: Elementary Charge
The charge of a single electron is approximately \( 1.6 \times 10^{-19} \, \text{C} \).
Step 2: Total Number of Electrons
To find the number of electrons in one coulomb of charge, we divide 1 coulomb by the charge of a single electron: \[ n = \frac{1 \, \text{C}}{1.6 \times 10^{-19} \, \text{C/electron}} = 6.25 \times 10^{18} \, \text{electrons} \]
Conclusion:
Therefore, one coulomb of charge contains approximately \( 6.25 \times 10^{18} \) electrons.
Step 1: Elementary Charge
The charge of a single electron is approximately \( 1.6 \times 10^{-19} \, \text{C} \).
Step 2: Total Number of Electrons
To find the number of electrons in one coulomb of charge, we divide 1 coulomb by the charge of a single electron: \[ n = \frac{1 \, \text{C}}{1.6 \times 10^{-19} \, \text{C/electron}} = 6.25 \times 10^{18} \, \text{electrons} \]
Conclusion:
Therefore, one coulomb of charge contains approximately \( 6.25 \times 10^{18} \) electrons.
8. The electric charge in uniform motion produces:
A. An electric field only
B. A magnetic field only
C. Both electric and magnetic field
D. Neither electric nor magnetic field
The correct answer is C. Both electric and magnetic field. Here's the explanation:
Step 1: Electric Field of a Charge
Any stationary electric charge produces an electric field in the space surrounding it.
Step 2: Motion of the Charge
When the charge is in uniform motion, in addition to the electric field, it also generates a magnetic field due to the movement of charge (current).
Conclusion:
Therefore, a charge in uniform motion produces both an electric field and a magnetic field.
Step 1: Electric Field of a Charge
Any stationary electric charge produces an electric field in the space surrounding it.
Step 2: Motion of the Charge
When the charge is in uniform motion, in addition to the electric field, it also generates a magnetic field due to the movement of charge (current).
Conclusion:
Therefore, a charge in uniform motion produces both an electric field and a magnetic field.
9. Identify the wrong statement:
A. Charge is a vector quantity
B. Current is a scalar quantity
C. Charge can be quantised
D. Charge is additive in nature
The correct answer is B. Current is a scalar quantity. Here's the explanation:
Step 1: Nature of Charge
Charge is indeed a scalar quantity, not a vector quantity. It has magnitude but no direction.
Step 2: Nature of Current
Current, however, is a scalar quantity. While it has direction associated with the flow of charge, the value itself is considered a scalar.
Step 3: Quantisation and Additivity
Charge is quantised (can only exist in integer multiples of the elementary charge) and is additive in nature (the total charge is the algebraic sum of individual charges).
Conclusion:
Therefore, the statement that charge is a vector quantity is incorrect.
Step 1: Nature of Charge
Charge is indeed a scalar quantity, not a vector quantity. It has magnitude but no direction.
Step 2: Nature of Current
Current, however, is a scalar quantity. While it has direction associated with the flow of charge, the value itself is considered a scalar.
Step 3: Quantisation and Additivity
Charge is quantised (can only exist in integer multiples of the elementary charge) and is additive in nature (the total charge is the algebraic sum of individual charges).
Conclusion:
Therefore, the statement that charge is a vector quantity is incorrect.
10. If a charge on the body is -1 nC, then how many excess electrons are present on the body?
A. \( 6.25 \times 10^{9} \) electrons
B. \( 1.6 \times 10^{9} \) electrons
C. \( 1.0 \times 10^{10} \) electrons
D. \( 5.0 \times 10^{9} \) electrons
The correct answer is A. \( 6.25 \times 10^{9} \) electrons. Here's the calculation:
Step 1: Charge of a Single Electron
The charge of a single electron is approximately \( 1.6 \times 10^{-19} \, \text{C} \).
Step 2: Total Charge
Given that the charge on the body is -1 nC (or \( -1 \times 10^{-9} \, \text{C} \)), we can find the number of excess electrons using the formula: \[ n = \frac{\text{Total Charge}}{\text{Charge of one electron}} = \frac{-1 \times 10^{-9} \, \text{C}}{-1.6 \times 10^{-19} \, \text{C/electron}} \]
Step 3: Calculation
\[ n = \frac{1 \times 10^{-9}}{1.6 \times 10^{-19}} \approx 6.25 \times 10^{9} \, \text{electrons} \]
Conclusion:
Therefore, the body has approximately \( 6.25 \times 10^{9} \) excess electrons.
Step 1: Charge of a Single Electron
The charge of a single electron is approximately \( 1.6 \times 10^{-19} \, \text{C} \).
Step 2: Total Charge
Given that the charge on the body is -1 nC (or \( -1 \times 10^{-9} \, \text{C} \)), we can find the number of excess electrons using the formula: \[ n = \frac{\text{Total Charge}}{\text{Charge of one electron}} = \frac{-1 \times 10^{-9} \, \text{C}}{-1.6 \times 10^{-19} \, \text{C/electron}} \]
Step 3: Calculation
\[ n = \frac{1 \times 10^{-9}}{1.6 \times 10^{-19}} \approx 6.25 \times 10^{9} \, \text{electrons} \]
Conclusion:
Therefore, the body has approximately \( 6.25 \times 10^{9} \) excess electrons.
11. A cylindrical conductor is placed near another positively charged conductor. The net charge acquired by the cylindrical conductor will be:
A. Positive only
B. Negative only
C. Zero
D. Either positive or negative
The correct answer is C. Zero. Here's the explanation:
Step 1: Induction Process
When a positively charged conductor is brought near a neutral conductor, it induces a separation of charges within the neutral conductor.
Step 2: Charge Distribution
The side of the cylindrical conductor closest to the positively charged conductor becomes negatively charged due to the attraction of electrons. The far side becomes positively charged due to a deficiency of electrons.
Step 3: Net Charge Acquired
The cylindrical conductor remains neutral overall (net charge zero) unless it is connected to a ground or another charge source. The distribution of charges leads to an effective polarization without a net charge.
Conclusion:
Therefore, the net charge acquired by the cylindrical conductor will be zero.
Step 1: Induction Process
When a positively charged conductor is brought near a neutral conductor, it induces a separation of charges within the neutral conductor.
Step 2: Charge Distribution
The side of the cylindrical conductor closest to the positively charged conductor becomes negatively charged due to the attraction of electrons. The far side becomes positively charged due to a deficiency of electrons.
Step 3: Net Charge Acquired
The cylindrical conductor remains neutral overall (net charge zero) unless it is connected to a ground or another charge source. The distribution of charges leads to an effective polarization without a net charge.
Conclusion:
Therefore, the net charge acquired by the cylindrical conductor will be zero.
12. Two point charges, +3 µC and -3 µC, are placed 10 cm apart. The electric field at the midpoint between them is:
A. Zero
B. 1.8 × 10^4 N/C
C. 3.6 × 10^4 N/C
D. 2.4 × 10^4 N/C
The correct answer is A. Zero. Here's the explanation:
The electric field produced by each charge at the midpoint cancels out because they are equal in magnitude but opposite in direction. Thus, the net electric field at the midpoint is zero.
The electric field produced by each charge at the midpoint cancels out because they are equal in magnitude but opposite in direction. Thus, the net electric field at the midpoint is zero.
13. If the distance between two charges is doubled, the force between them becomes:
A. One-fourth
B. Half
C. Double
D. Four times
The correct answer is A. One-fourth. Here's the explanation:
According to Coulomb's law, the force between two charges is inversely proportional to the square of the distance between them. Therefore, if the distance is doubled, the force becomes one-fourth.
According to Coulomb's law, the force between two charges is inversely proportional to the square of the distance between them. Therefore, if the distance is doubled, the force becomes one-fourth.
14. The electric potential energy of a system of two point charges is given by:
A. \( \frac{q_1 q_2}{r} \)
B. \( k \frac{q_1 q_2}{r} \)
C. \( \frac{1}{2} k \frac{q_1 q_2}{r^2} \)
D. \( k \frac{q_1 + q_2}{r} \)
The correct answer is B. \( k \frac{q_1 q_2}{r} \). Here's the explanation:
The electric potential energy \( U \) of two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by \( U = k \frac{q_1 q_2}{r} \), where \( k \) is Coulomb's constant.
The electric potential energy \( U \) of two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by \( U = k \frac{q_1 q_2}{r} \), where \( k \) is Coulomb's constant.
15. The capacitance of a parallel plate capacitor depends on:
A. Area of the plates and distance between them
B. Only the area of the plates
C. Only the distance between the plates
D. The type of material between the plates
The correct answer is A. Area of the plates and distance between them. Here's the explanation:
The capacitance \( C \) of a parallel plate capacitor is given by \( C = \frac{\varepsilon A}{d} \), where \( \varepsilon \) is the permittivity of the dielectric between the plates, \( A \) is the area of the plates, and \( d \) is the distance between them. Thus, it depends on both the area and the distance.
The capacitance \( C \) of a parallel plate capacitor is given by \( C = \frac{\varepsilon A}{d} \), where \( \varepsilon \) is the permittivity of the dielectric between the plates, \( A \) is the area of the plates, and \( d \) is the distance between them. Thus, it depends on both the area and the distance.
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