Test Your Skills: MCQs on Electrostatic Potential and Capacitance!
Time Left: 45:00
Total Marks: 60, Obtained Marks: 0
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 wire is bent in a circle of radius 10 cm. It is given a charge of 200 µC, which spreads on it uniformly. Calculate the electric potential at its center.
A. \( 1.2 \times 10^5 \, \text{V} \)
B. \( 1.5 \times 10^5 \, \text{V} \)
C. \( 1.8 \times 10^6 \, \text{V} \)
D. \( 2.0 \times 10^5 \, \text{V} \)
The correct answer is C. \( 1.8 \times 10^5 \, \text{V} \). Here's how it works:
Step 1: Formula for Electric Potential
The potential at the center of a uniformly charged circular wire is given by the formula:
\[ V = \frac{kQ}{R} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( Q = 200 \, \mu\text{C} \), and \( R = 0.1 \, \text{m} \).
Step 2: Substituting the Values
Substituting the known values into the formula: \[ V = \frac{9 \times 10^9 \times 200 \times 10^{-6}}{0.1} = 1.8 \times 10^6 \, \text{V} \]
Conclusion:
The electric potential at the center of the circular wire is \( 1.8 \times 10^6 \, \text{V} \).
Step 1: Formula for Electric Potential
The potential at the center of a uniformly charged circular wire is given by the formula:
\[ V = \frac{kQ}{R} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( Q = 200 \, \mu\text{C} \), and \( R = 0.1 \, \text{m} \).
Step 2: Substituting the Values
Substituting the known values into the formula: \[ V = \frac{9 \times 10^9 \times 200 \times 10^{-6}}{0.1} = 1.8 \times 10^6 \, \text{V} \]
Conclusion:
The electric potential at the center of the circular wire is \( 1.8 \times 10^6 \, \text{V} \).
3. At a certain distance from a point charge, the electric field is 500 V/m and the potential is 3000 V. What is the distance?
A. 3 m
B. 4 m
C. 6 m
D. 9 m
The correct answer is C. 6 m. Here's how it works:
Step 1: Formula for Electric Field and Potential
The electric field due to a point charge is given by: \[ E = \frac{kQ}{r^2} \] and the potential is: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( E = 500 \, \text{V/m} \), and \( V = 3000 \, \text{V} \).
Step 2: Solving for Distance
By dividing the potential by the electric field: \[ r = \frac{V}{E} = \frac{3000}{500} = 6 \, \text{m} \]
Conclusion:
The distance from the point charge is \( 6 \, \text{m} \).
Step 1: Formula for Electric Field and Potential
The electric field due to a point charge is given by: \[ E = \frac{kQ}{r^2} \] and the potential is: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( E = 500 \, \text{V/m} \), and \( V = 3000 \, \text{V} \).
Step 2: Solving for Distance
By dividing the potential by the electric field: \[ r = \frac{V}{E} = \frac{3000}{500} = 6 \, \text{m} \]
Conclusion:
The distance from the point charge is \( 6 \, \text{m} \).
4. Electric charges of + 10 µC, + 5 µC, - 3 µC, and + 8 µC are placed at the corners of a square of side \( \sqrt{2} \, \text{m} \). What is the potential at the center of the square?
A. 100,000 V
B. 150,000 V
C. 180,000 V
D. 200,000 V
The correct answer is C. 180,000 V. Here's how it works:
Step 1: Formula for Electric Potential
The potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), and the distance from each corner to the center is 1 m.
Step 2: Summing Potentials
The total potential is the sum of the potentials due to each charge: \[ V_{\text{total}} = k \times \left( 10 \times 10^{-6} + 5 \times 10^{-6} - 3 \times 10^{-6} + 8 \times 10^{-6} \right) \] Simplifying: \[ V_{\text{total}} = 9 \times 10^9 \times 20 \times 10^{-6} = 180,000 \, \text{V} \]
Conclusion:
The potential at the center of the square is \( 180,000 \, \text{V} \).
Step 1: Formula for Electric Potential
The potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), and the distance from each corner to the center is 1 m.
Step 2: Summing Potentials
The total potential is the sum of the potentials due to each charge: \[ V_{\text{total}} = k \times \left( 10 \times 10^{-6} + 5 \times 10^{-6} - 3 \times 10^{-6} + 8 \times 10^{-6} \right) \] Simplifying: \[ V_{\text{total}} = 9 \times 10^9 \times 20 \times 10^{-6} = 180,000 \, \text{V} \]
Conclusion:
The potential at the center of the square is \( 180,000 \, \text{V} \).
5. The potential at a point due to a positive charge of 100 µC at a distance of 9 m is:
A. 1,000,000 V
B. 100,000 V
C. 1,200,000 V
D. 750,000 V
The correct answer is B. 100,000 V. Here's how it works:
Step 1: Formula for Electric Potential
The potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( Q = 100 \times 10^{-6} \, \text{C} \), and \( r = 9 \, \text{m} \).
Step 2: Substituting Values
Substituting the values into the formula: \[ V = \frac{9 \times 10^9 \times 100 \times 10^{-6}}{9} = 100,000 \, \text{V} \]
Conclusion:
The potential at the point is \( 100,000 \, \text{V} \).
Step 1: Formula for Electric Potential
The potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( Q = 100 \times 10^{-6} \, \text{C} \), and \( r = 9 \, \text{m} \).
Step 2: Substituting Values
Substituting the values into the formula: \[ V = \frac{9 \times 10^9 \times 100 \times 10^{-6}}{9} = 100,000 \, \text{V} \]
Conclusion:
The potential at the point is \( 100,000 \, \text{V} \).
6. A charge of 50 µC is placed at a distance of 5 m from a point. What is the electric potential at that point due to the charge?
A. 45,000 V
B. 90,000 V
C. 75,000 V
D. 120,000 V
The correct answer is B. 90,000 V. Here's how it works:
Step 1: Formula for Electric Potential
The potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( Q = 50 \times 10^{-6} \, \text{C} \), and \( r = 5 \, \text{m} \).
Step 2: Substituting Values
Substituting the values into the formula: \[ V = \frac{9 \times 10^9 \times 50 \times 10^{-6}}{5} = 90,000 \, \text{V} \]
Conclusion:
The potential at the point is \( 90,000 \, \text{V} \).
Step 1: Formula for Electric Potential
The potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( Q = 50 \times 10^{-6} \, \text{C} \), and \( r = 5 \, \text{m} \).
Step 2: Substituting Values
Substituting the values into the formula: \[ V = \frac{9 \times 10^9 \times 50 \times 10^{-6}}{5} = 90,000 \, \text{V} \]
Conclusion:
The potential at the point is \( 90,000 \, \text{V} \).
7. The electric potential at 0.9 m from a point charge is +50 V. What is the magnitude and sign of the charge?
A. +5 nC
B. -5 nC
C. +10 nC
D. -10 nC
The correct answer is A. +5 nC. Here's how it works:
Step 1: Formula for Electric Potential
The potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( V = +50 \, \text{V} \), \( r = 0.9 \, \text{m} \), and \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
Step 2: Solve for Charge \( Q \)
Rearranging the formula to solve for \( Q \): \[ Q = \frac{V \cdot r}{k} = \frac{50 \times 0.9}{9 \times 10^9} = 5 \times 10^{-9} \, \text{C} = 5 \, \text{nC} \]
Conclusion:
The charge has a magnitude of \( 5 \, \text{nC} \) and is positive.
Step 1: Formula for Electric Potential
The potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( V = +50 \, \text{V} \), \( r = 0.9 \, \text{m} \), and \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
Step 2: Solve for Charge \( Q \)
Rearranging the formula to solve for \( Q \): \[ Q = \frac{V \cdot r}{k} = \frac{50 \times 0.9}{9 \times 10^9} = 5 \times 10^{-9} \, \text{C} = 5 \, \text{nC} \]
Conclusion:
The charge has a magnitude of \( 5 \, \text{nC} \) and is positive.
8. The electric field at a point due to a point charge is 20 N/C and the electric potential at that point is 10 J/C. Calculate the distance of the point from the charge and the magnitude of the charge.
A. Distance = 1 m, Charge = 1 nC
B. Distance = 0.5 m, Charge = 0.555 nC
C. Distance = 0.3 m, Charge = 2 nC
D. Distance = 2 m, Charge = 0.1 nC
The correct answer is B. Distance = 0.5 m, Charge = 0.555 nC. Here's how it works:
Step 1: Relationship between Electric Field and Potential
By dividing the formula for the electric field by the formula for electric potential, we can calculate the distance: \[ r = \frac{V}{E} = \frac{10}{20} = 0.5 \, \text{m} \]
Step 2: Calculate Charge
Using the formula for potential \( V = \frac{kQ}{r} \), solve for \( Q \): \[ Q = \frac{Vr}{k} = \frac{10 \times 0.5}{9 \times 10^9} = 0.555 \, \text{nC} \]
Conclusion:
The distance is \( 0.5 \, \text{m} \) and the charge is \( 0.555 \, \text{nC} \).
Step 1: Relationship between Electric Field and Potential
By dividing the formula for the electric field by the formula for electric potential, we can calculate the distance: \[ r = \frac{V}{E} = \frac{10}{20} = 0.5 \, \text{m} \]
Step 2: Calculate Charge
Using the formula for potential \( V = \frac{kQ}{r} \), solve for \( Q \): \[ Q = \frac{Vr}{k} = \frac{10 \times 0.5}{9 \times 10^9} = 0.555 \, \text{nC} \]
Conclusion:
The distance is \( 0.5 \, \text{m} \) and the charge is \( 0.555 \, \text{nC} \).
9. A point charge of +15 µC is placed at the origin. Calculate the electric potential at a distance of 4 m from the charge.
A. 33750 V
B. 15000 V
C. 10000 V
D. 20000 V
The correct answer is A. 33750 V. Here's how it works:
Step 1: Formula for Electric Potential
The electric potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( Q = 15 \times 10^{-6} \, \text{C} \), and \( r = 4 \, \text{m} \).
Step 2: Substitute Values
\[ V = \frac{9 \times 10^9 \times 15 \times 10^{-6}}{4} = 33750 \, \text{V} \]
Conclusion:
The electric potential at a distance of 4 m from the charge is \( 33750 \, V \).
Step 1: Formula for Electric Potential
The electric potential due to a point charge is given by: \[ V = \frac{kQ}{r} \] where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( Q = 15 \times 10^{-6} \, \text{C} \), and \( r = 4 \, \text{m} \).
Step 2: Substitute Values
\[ V = \frac{9 \times 10^9 \times 15 \times 10^{-6}}{4} = 33750 \, \text{V} \]
Conclusion:
The electric potential at a distance of 4 m from the charge is \( 33750 \, V \).
10. Two charges 3 x 10-8 C and -2 x 10-8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
A. 6 cm from the negative charge
B. 9 cm from the positive charge
C. 10 cm from the negative charge
D. 7.5 cm from the center
The correct answer is A. 9 cm from the positive charge. Here's how it works:
Step 1: Setting the Total Potential to Zero
The total potential at distance \( x \) from the positive charge and \( 0.15 - x \) from the negative charge is given by: \[ \frac{3 \times 10^{-8}}{x} - \frac{2 \times 10^{-8}}{0.15 - x} = 0 \]
Step 2: Rearranging and Solving for \( x \)
\[ 3 \times 10^{-8} (0.15 - x) = 2 \times 10^{-8} x \]
Solving the equation gives: \[ x = 0.09 \, m = 9 \, cm \]
Conclusion:
The electric potential is zero at a point 9 cm from the positive charge.
Step 1: Setting the Total Potential to Zero
The total potential at distance \( x \) from the positive charge and \( 0.15 - x \) from the negative charge is given by: \[ \frac{3 \times 10^{-8}}{x} - \frac{2 \times 10^{-8}}{0.15 - x} = 0 \]
Step 2: Rearranging and Solving for \( x \)
\[ 3 \times 10^{-8} (0.15 - x) = 2 \times 10^{-8} x \]
Solving the equation gives: \[ x = 0.09 \, m = 9 \, cm \]
Conclusion:
The electric potential is zero at a point 9 cm from the positive charge.
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. 10.55 nF
B. 22.13 nF
C. 44.25 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 electric potential at a point due to two point charges,
\( +4 \, \mu C \) and \( -2 \, \mu C \), separated by a distance of \( 0.5 \, m \).
The point of interest is located \( 0.2 \, m \) from the \( +4 \, \mu C \) charge
along the line connecting the two charges.
A. 60 V
B. 45 V
C. 30 V
D. 15 V
The correct answer is A. 60 V. Here's the explanation:
Step 1: Identify Distances
The distances to the point of interest are: - From \( +4 \, \mu C \): \( r_1 = 0.2 \, m \) - From \( -2 \, \mu C \): \( r_2 = 0.5 \, m - 0.2 \, m = 0.3 \, m \)
Step 2: Calculate Electric Potential
Using the formula \( V = k \cdot \frac{Q}{r} \):
For \( +4 \, \mu C \): \[ V_1 = 9 \times 10^9 \cdot \frac{4 \times 10^{-6}}{0.2} = 180000 \, V \]
For \( -2 \, \mu C \): \[ V_2 = 9 \times 10^9 \cdot \frac{-2 \times 10^{-6}}{0.3} \approx -60000 \, V \]
Step 3: Total Electric Potential
\[ V_{\text{total}} = V_1 + V_2 = 180000 - 60000 = 120000 \, V \]
Conclusion:
Therefore, the total electric potential at the point is approximately \( 120000 \, V \).
Step 1: Identify Distances
The distances to the point of interest are: - From \( +4 \, \mu C \): \( r_1 = 0.2 \, m \) - From \( -2 \, \mu C \): \( r_2 = 0.5 \, m - 0.2 \, m = 0.3 \, m \)
Step 2: Calculate Electric Potential
Using the formula \( V = k \cdot \frac{Q}{r} \):
For \( +4 \, \mu C \): \[ V_1 = 9 \times 10^9 \cdot \frac{4 \times 10^{-6}}{0.2} = 180000 \, V \]
For \( -2 \, \mu C \): \[ V_2 = 9 \times 10^9 \cdot \frac{-2 \times 10^{-6}}{0.3} \approx -60000 \, V \]
Step 3: Total Electric Potential
\[ V_{\text{total}} = V_1 + V_2 = 180000 - 60000 = 120000 \, V \]
Conclusion:
Therefore, the total electric potential at the point is approximately \( 120000 \, V \).
14. Two positive point charges of \(12 \, \mu C\) and \(8 \, \mu C\)
are \(10 \, cm\) apart. The work done in bringing them \(4 \, cm\) closer is:
A. 4.5 J
B. 5.754 J
C. 7.2 J
D. 6.0 J
The correct answer is A. 5.754 J. Here's the explanation:
Step 1: Identify Initial and Final Distances
Initial distance \( r_1 = 0.1 \, m \), Final distance \( r_2 = 0.06 \, m \).
Step 2: Calculate Initial Potential Energy
Using the formula \( U_1 = k \cdot \frac{Q_1 \cdot Q_2}{r_1} \):
\( U_1 = 8.99 \times 10^9 \cdot \frac{(12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.1} \approx 8.63 \, J \).
Step 3: Calculate Final Potential Energy
\( U_2 = 8.99 \times 10^9 \cdot \frac{(12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.06} \approx 14.384 \, J \).
Step 4: Calculate Work Done
\( W = U_2 - U_1 = 14.384 - 8.63 \approx 5.754 \, J \).
Conclusion:
Therefore, the work done in bringing the charges \(4 \, cm\) closer is approximately \(5.754 \, J\).
Step 1: Identify Initial and Final Distances
Initial distance \( r_1 = 0.1 \, m \), Final distance \( r_2 = 0.06 \, m \).
Step 2: Calculate Initial Potential Energy
Using the formula \( U_1 = k \cdot \frac{Q_1 \cdot Q_2}{r_1} \):
\( U_1 = 8.99 \times 10^9 \cdot \frac{(12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.1} \approx 8.63 \, J \).
Step 3: Calculate Final Potential Energy
\( U_2 = 8.99 \times 10^9 \cdot \frac{(12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.06} \approx 14.384 \, J \).
Step 4: Calculate Work Done
\( W = U_2 - U_1 = 14.384 - 8.63 \approx 5.754 \, J \).
Conclusion:
Therefore, the work done in bringing the charges \(4 \, cm\) closer is approximately \(5.754 \, J\).
15. Ten electrons are equally spaced and fixed around a circle of radius \( R \).
Relative to \( V = 0 \) at infinity, what are the electrostatic potential \( V \)
and the electric field \( E \) at the center \( C \)?
A. \( V = \frac{-16 \times 10^{-19} \cdot k}{R}, \, E = 0 \)
B. \( V = 0, \, E = 0 \)
C. \( V = \frac{16 \times 10^{-19} \cdot k}{R}, \, E = 0 \)
D. \( V = \frac{-16 \times 10^{-19} \cdot k}{R}, \, E \neq 0 \)
The correct answer is A. \( V = \frac{-16 \times 10^{-19} \cdot k}{R}, \, E = 0 \). Here's the explanation:
Step 1: Calculate the Electric Potential \( V \)
The total potential at the center due to the 10 electrons is given by:
\( V = 10 \cdot k \cdot \frac{-1.6 \times 10^{-19}}{R} = \frac{-16 \times 10^{-19} \cdot k}{R} \).
Step 2: Calculate the Electric Field \( E \)
Due to the symmetry of the arrangement, the electric field contributions from all electrons cancel out at the center:
\( E = 0 \).
Conclusion:
Therefore, the electrostatic potential at the center \( C \) is \( V = \frac{-16 \times 10^{-19} \cdot k}{R} \), and the electric field \( E \) is zero.
Step 1: Calculate the Electric Potential \( V \)
The total potential at the center due to the 10 electrons is given by:
\( V = 10 \cdot k \cdot \frac{-1.6 \times 10^{-19}}{R} = \frac{-16 \times 10^{-19} \cdot k}{R} \).
Step 2: Calculate the Electric Field \( E \)
Due to the symmetry of the arrangement, the electric field contributions from all electrons cancel out at the center:
\( E = 0 \).
Conclusion:
Therefore, the electrostatic potential at the center \( C \) is \( V = \frac{-16 \times 10^{-19} \cdot k}{R} \), and the electric field \( E \) is zero.
.png)
0 Comments