"Capacitors MCQ Test: Assess Your Understanding of Capacitance and Energy!"
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1. No current flows between two charged bodies connected together when they have the same:
A. Capacitance
B. Charge
C. Resistance
D. Potential
The correct answer is D. Potential. Here's how it works:
Step 1: Current Flow
Current flows when there is a potential difference between two bodies. This potential difference causes charge movement.
Step 2: Equal Potentials
When the potentials of two connected bodies are the same, there is no potential difference, and hence, no current flows.
Conclusion:
No current flows if both bodies have the same potential.
Step 1: Current Flow
Current flows when there is a potential difference between two bodies. This potential difference causes charge movement.
Step 2: Equal Potentials
When the potentials of two connected bodies are the same, there is no potential difference, and hence, no current flows.
Conclusion:
No current flows if both bodies have the same potential.
2. A metal rod is moved through a magnetic field. The induced emf in the rod is:
A. Proportional to the resistance of the rod
B. Independent of the speed of the rod
C. Proportional to the speed of the rod
D. Zero if the rod is moving perpendicular to the field
The correct answer is C. Proportional to the speed of the rod. Here's how it works:
Step 1: Faraday's Law of Electromagnetic Induction
According to Faraday's Law, the induced emf is proportional to the rate of change of magnetic flux, which depends on the speed of the rod moving through the magnetic field.
Step 2: Motion Through the Magnetic Field
When the rod moves faster, the rate of change of magnetic flux increases, thereby increasing the induced emf.
Conclusion:
The induced emf is proportional to the speed of the rod as it moves through the magnetic field.
Step 1: Faraday's Law of Electromagnetic Induction
According to Faraday's Law, the induced emf is proportional to the rate of change of magnetic flux, which depends on the speed of the rod moving through the magnetic field.
Step 2: Motion Through the Magnetic Field
When the rod moves faster, the rate of change of magnetic flux increases, thereby increasing the induced emf.
Conclusion:
The induced emf is proportional to the speed of the rod as it moves through the magnetic field.
3. The capacity of a parallel plate condenser is \( C \). Its capacity when the separation between the plates is halved will be:
A. \( C/2 \)
B. \( 2C \)
C. \( 4C \)
D. \( C \)
The correct answer is B. \( 2C \). Here's the explanation:
Step 1: Formula for Capacitance
The capacitance of a parallel plate capacitor is given by \( C = \frac{\varepsilon_0 A}{d} \), where \( A \) is the area of the plates, \( d \) is the separation between the plates, and \( \varepsilon_0 \) is the permittivity of free space.
Step 2: Effect of Plate Separation
When the separation \( d \) is halved, the new capacitance becomes \( C' = \frac{\varepsilon_0 A}{d/2} = 2C \).
Conclusion:
The capacitance doubles when the separation between the plates is halved.
Step 1: Formula for Capacitance
The capacitance of a parallel plate capacitor is given by \( C = \frac{\varepsilon_0 A}{d} \), where \( A \) is the area of the plates, \( d \) is the separation between the plates, and \( \varepsilon_0 \) is the permittivity of free space.
Step 2: Effect of Plate Separation
When the separation \( d \) is halved, the new capacitance becomes \( C' = \frac{\varepsilon_0 A}{d/2} = 2C \).
Conclusion:
The capacitance doubles when the separation between the plates is halved.
4. The energy of a charged capacitor is given by the expression (where \( q \) is the charge on the conductor and \( C \) is its capacitance):
A. \( E = \frac{q^2}{2} \)
B. \( E = \frac{q^2}{2C} \)
C. \( E = \frac{2q^2}{C} \)
D. \( E = \frac{C^2}{2q} \)
The correct answer is B. \( E = \frac{q^2}{2C} \). Here's the explanation:
Step 1: Energy stored in a capacitor
The energy stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \), where \( V \) is the potential difference across the capacitor.
Step 2: Substituting the relationship between charge and voltage
Since \( q = C \cdot V \), we can substitute \( V = \frac{q}{C} \) into the energy equation, giving:
\( E = \frac{1}{2} C \left(\frac{q}{C}\right)^2 = \frac{q^2}{2C} \).
Conclusion:
The energy stored in a charged capacitor is \( E = \frac{q^2}{2C} \).
Step 1: Energy stored in a capacitor
The energy stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \), where \( V \) is the potential difference across the capacitor.
Step 2: Substituting the relationship between charge and voltage
Since \( q = C \cdot V \), we can substitute \( V = \frac{q}{C} \) into the energy equation, giving:
\( E = \frac{1}{2} C \left(\frac{q}{C}\right)^2 = \frac{q^2}{2C} \).
Conclusion:
The energy stored in a charged capacitor is \( E = \frac{q^2}{2C} \).
5. The energy stored in a capacitor with capacitance \( C \), raised to a potential \( V \), is given by:
A. \( E = \frac{C^2 V}{2} \)
B. \( E = \frac{1}{2} C V^2 \)
C. \( E = \frac{C V}{2} \)
D. \( E = C V \)
The correct answer is B. \( E = \frac{1}{2} C V^2 \). Here's the explanation:
Step 1: Energy stored in a capacitor
The energy stored in a capacitor is given by the formula \( E = \frac{1}{2} C V^2 \), where:
\( C \) is the capacitance and \( V \) is the potential difference across the plates of the capacitor.
Step 2: Understanding the formula
This formula comes from the integration of the work required to charge the capacitor. It shows that the energy stored in the capacitor is proportional to both the capacitance \( C \) and the square of the potential \( V \).
Conclusion:
The energy stored in a capacitor is \( E = \frac{1}{2} C V^2 \).
Step 1: Energy stored in a capacitor
The energy stored in a capacitor is given by the formula \( E = \frac{1}{2} C V^2 \), where:
\( C \) is the capacitance and \( V \) is the potential difference across the plates of the capacitor.
Step 2: Understanding the formula
This formula comes from the integration of the work required to charge the capacitor. It shows that the energy stored in the capacitor is proportional to both the capacitance \( C \) and the square of the potential \( V \).
Conclusion:
The energy stored in a capacitor is \( E = \frac{1}{2} C V^2 \).
6. 64 drops, each having capacitance \( C \) and potential \( V \), are combined to form a big drop. If the charge on the small drop is \( q \), then the charge on the big drop will be:
A. \( 16q \)
B. \( 64q \)
C. \( 128q \)
D. \( 32q \)
The correct answer is B. \( 64q \). Here's the explanation:
Step 1: Charge of individual drops
The charge on a small drop is \( q \), and when 64 drops combine, their charges add up.
Step 2: Total charge on the big drop
Since the charges of all 64 small drops combine, the total charge on the big drop will be \( 64q \).
Conclusion:
The charge on the big drop will be \( 64q \).
Step 1: Charge of individual drops
The charge on a small drop is \( q \), and when 64 drops combine, their charges add up.
Step 2: Total charge on the big drop
Since the charges of all 64 small drops combine, the total charge on the big drop will be \( 64q \).
Conclusion:
The charge on the big drop will be \( 64q \).
7. A parallel plate capacitor has a capacitance \( C \) when the distance between the plates is \( d \) and the area of the plates is \( A \). If the distance between the plates is doubled, what will be the new capacitance?
A. \( \frac{C}{2} \)
B. \( 2C \)
C. \( C \)
D. \( 4C \)
The correct answer is A. \( \frac{C}{2} \). Here's the explanation:
Step 1: Capacitance Formula
The capacitance \( C \) of a parallel plate capacitor is given by the formula: \[ C = \frac{\epsilon_0 A}{d} \] where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between them.
Step 2: Effect of Doubling Distance
If the distance \( d \) is doubled, the new capacitance \( C' \) becomes: \[ C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2} \]
Conclusion:
Therefore, when the distance between the plates is doubled, the new capacitance is \( \frac{C}{2} \).
Step 1: Capacitance Formula
The capacitance \( C \) of a parallel plate capacitor is given by the formula: \[ C = \frac{\epsilon_0 A}{d} \] where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between them.
Step 2: Effect of Doubling Distance
If the distance \( d \) is doubled, the new capacitance \( C' \) becomes: \[ C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2} \]
Conclusion:
Therefore, when the distance between the plates is doubled, the new capacitance is \( \frac{C}{2} \).
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. The force acting upon a charged particle kept between the plates of a charged condenser is \( F \). If one plate of the condenser is removed, then the force acting on the same particle will become:
A. 0
B. \( \frac{F}{2} \)
C. \( F \)
D. \( 2F \)
The correct answer is B. \( \frac{F}{2} \). Here's the explanation:
Step 1: Understanding the Force Between Plates
A charged particle experiences a force due to the electric field between the plates of a charged condenser.
Step 2: Effect of Removing One Plate
If one plate is removed, the remaining plate creates an electric field, but the field will not be uniform. The force acting on the charged particle will be halved.
Conclusion:
Therefore, the force acting on the charged particle becomes \( \frac{F}{2} \) when one plate is removed.
Step 1: Understanding the Force Between Plates
A charged particle experiences a force due to the electric field between the plates of a charged condenser.
Step 2: Effect of Removing One Plate
If one plate is removed, the remaining plate creates an electric field, but the field will not be uniform. The force acting on the charged particle will be halved.
Conclusion:
Therefore, the force acting on the charged particle becomes \( \frac{F}{2} \) when one plate is removed.
10. The force between the plates of a parallel plate capacitor of capacitance \( C \) and distance \( d \) of separation with a potential difference \( V \) between the plates is:
A. \( \frac{C \cdot V^2}{d} \)
B. \( \frac{C \cdot V}{d} \)
C. \( \frac{1}{2} \frac{C \cdot V^2}{d} \)
D. \( C \cdot V^2 \)
The correct answer is A. \( \frac{C \cdot V^2}{d} \). Here's the explanation:
Step 1: Electric Field
The electric field \( E \) between the plates is given by \( E = \frac{V}{d} \).
Step 2: Charge on the Plates
The charge \( Q \) on the plates is \( Q = C \cdot V \).
Step 3: Force Calculation
The force \( F \) on one plate is calculated as \( F = Q \cdot E = (C \cdot V) \cdot \left(\frac{V}{d}\right) = \frac{C \cdot V^2}{d} \).
Conclusion:
Therefore, the force between the plates of the capacitor is given by \( F = \frac{C \cdot V^2}{d} \).
Step 1: Electric Field
The electric field \( E \) between the plates is given by \( E = \frac{V}{d} \).
Step 2: Charge on the Plates
The charge \( Q \) on the plates is \( Q = C \cdot V \).
Step 3: Force Calculation
The force \( F \) on one plate is calculated as \( F = Q \cdot E = (C \cdot V) \cdot \left(\frac{V}{d}\right) = \frac{C \cdot V^2}{d} \).
Conclusion:
Therefore, the force between the plates of the capacitor is given by \( F = \frac{C \cdot V^2}{d} \).
11. The capacity of a parallel plate condenser is 15 µF when the distance between its plates is 6 cm. If the distance between the plates is reduced to 2 cm, then the capacity of this parallel plates condenser will be:
A. 45 µF
B. 30 µF
C. 15 µF
D. 60 µF
The correct answer is A. 45 µF. Here's the explanation:
Step 1: Understanding Capacitance
The capacitance of a parallel plate capacitor is given by the formula \( C = \frac{\varepsilon_0 \cdot A}{d} \).
Step 2: Ratio of Capacitances
Since the area \( A \) and the permittivity \( \varepsilon_0 \) remain constant, we have \( \frac{C_1}{C_2} = \frac{d_2}{d_1} \).
Step 3: Calculate New Capacitance
Substituting the values, we find \( C_2 = C_1 \cdot \frac{d_1}{d_2} = 15 \, \mu F \cdot \frac{6 \, cm}{2 \, cm} = 45 \, \mu F \).
Conclusion:
Therefore, the new capacitance when the distance is reduced to 2 cm is \( 45 \, \mu F \).
Step 1: Understanding Capacitance
The capacitance of a parallel plate capacitor is given by the formula \( C = \frac{\varepsilon_0 \cdot A}{d} \).
Step 2: Ratio of Capacitances
Since the area \( A \) and the permittivity \( \varepsilon_0 \) remain constant, we have \( \frac{C_1}{C_2} = \frac{d_2}{d_1} \).
Step 3: Calculate New Capacitance
Substituting the values, we find \( C_2 = C_1 \cdot \frac{d_1}{d_2} = 15 \, \mu F \cdot \frac{6 \, cm}{2 \, cm} = 45 \, \mu F \).
Conclusion:
Therefore, the new capacitance when the distance is reduced to 2 cm is \( 45 \, \mu F \).
12. A parallel plate capacitor has a plate area of 0.5 m² and a plate separation of 1 cm. What is its capacitance if the permittivity of free space is 8.85 × 10⁻¹² F/m?
A. 44.25 nF
B. 22.13 nF
C. 10.55 nF
D. 55.25 nF
The correct answer is A. 44.25 nF. Using the formula:
Capacitance \( C \) = \( \frac{\varepsilon_0 \cdot A}{d} \)
Substituting the values:
\( C = \frac{8.85 \times 10^{-12} \times 0.5}{0.01} = 44.25 \times 10^{-9} \, F = 44.25 \, nF \).
Capacitance \( C \) = \( \frac{\varepsilon_0 \cdot A}{d} \)
Substituting the values:
\( C = \frac{8.85 \times 10^{-12} \times 0.5}{0.01} = 44.25 \times 10^{-9} \, F = 44.25 \, nF \).
13. Calculate the energy stored in a capacitor with a capacitance of 10 µF charged to a voltage of 50 V.
A. 12.5 mJ
B. 25 mJ
C. 5 mJ
D. 10 mJ
The correct answer is A. 12.5 mJ. Using the formula:
Energy \( U \) = \( \frac{1}{2} C V^2 \)
Substituting the values:
\( U = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot (50)^2 = 12.5 \times 10^{-3} \, J = 12.5 \, mJ \).
Energy \( U \) = \( \frac{1}{2} C V^2 \)
Substituting the values:
\( U = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot (50)^2 = 12.5 \times 10^{-3} \, J = 12.5 \, mJ \).
14. A capacitor has a capacitance of 20 µF and is connected to a 100 V battery. What is the charge stored in the capacitor?
A. 2 mC
B. 0.5 mC
C. 1 mC
D. 0.2 mC
The correct answer is A. 2 mC. Using the formula:
Charge \( Q = C \cdot V \)
Substituting the values:
\( Q = 20 \times 10^{-6} \cdot 100 = 2 \times 10^{-3} \, C = 2 \, mC \).
Charge \( Q = C \cdot V \)
Substituting the values:
\( Q = 20 \times 10^{-6} \cdot 100 = 2 \times 10^{-3} \, C = 2 \, mC \).
15. The capacitance of a parallel plate capacitor is 30 µF with a plate separation of 4 cm. If the plate separation is reduced to 1 cm, what will be the new capacitance?
A. 120 µF
B. 60 µF
C. 30 µF
D. 90 µF
The correct answer is A. 120 µF. Using the capacitance ratio:
Capacitance \( C \) is inversely proportional to distance \( d \).
\( C_2 = C_1 \cdot \frac{d_1}{d_2} = 30 \cdot \frac{4}{1} = 120 \, \mu F \).
Capacitance \( C \) is inversely proportional to distance \( d \).
\( C_2 = C_1 \cdot \frac{d_1}{d_2} = 30 \cdot \frac{4}{1} = 120 \, \mu F \).
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