Electrostatic Potential And Capacitance

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| By Tanmay Shankar
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Tanmay Shankar
Community Contributor
Quizzes Created: 491 | Total Attempts: 1,782,797
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Electrostatic Potential And Capacitance - Quiz

Time: 30 Minute


Questions and Answers
  • 1. 

    A parallel plate capacitor has a capacitance of 50 pf in air and 105 pf when immersed in oil. The dielectric constant of the oil is:

    • A.

      50/105

    • B.

      1

    • C.

      2.1

    • D.
    Correct Answer
    C. 2.1
    Explanation
    The dielectric constant of a material is a measure of its ability to store electrical energy in an electric field. The capacitance of a parallel plate capacitor is directly proportional to the dielectric constant of the material between the plates. In this case, the capacitance of the capacitor increases from 50 pf to 105 pf when immersed in oil. By using the formula for capacitance, C = (k * ε₀ * A) / d, where k is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates, we can find that the dielectric constant of the oil is 2.1.

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  • 2. 

    Three capacitors of capacitance 4µ F, 6µF and 12µF are connected first in series and then in parallel. What is the ratio of equivalent capacitance in the two cases?

    • A.

      2 : 3

    • B.

      1 : 11

    • C.

      11 : 1

    • D.

      1 : 3

    Correct Answer
    B. 1 : 11
    Explanation
    When capacitors are connected in series, the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. So, for the given capacitors of capacitance 4µF, 6µF, and 12µF, the equivalent capacitance in series would be 1/(1/4 + 1/6 + 1/12) = 1/((3+2+1)/12) = 12/6 = 2µF.

    When capacitors are connected in parallel, the equivalent capacitance is the sum of the individual capacitances. So, for the given capacitors, the equivalent capacitance in parallel would be 4µF + 6µF + 12µF = 22µF.

    Therefore, the ratio of the equivalent capacitance in series to the equivalent capacitance in parallel is 2µF : 22µF, which simplifies to 1 : 11.

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  • 3. 

    A work of 100 joule is performed in carrying a charge of -5 coulomb from infinity to a particular point in an electrostatic field. The potential of this point is:

    • A.

      100 V

    • B.

      5 V

    • C.

      -20 V

    • D.

      20 V

    Correct Answer
    C. -20 V
    Explanation
    The work done in carrying a charge from infinity to a particular point in an electrostatic field is equal to the change in potential energy of the charge. Since the work done is positive (100 joules), it means that the potential energy of the charge has increased. In an electrostatic field, the potential decreases as we move away from the source of the field. Therefore, the potential at the particular point must be lower than infinity. The only option that satisfies this condition is -20 V, which indicates a potential 20 volts lower than infinity.

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  • 4. 

    When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance:

    • A.

      Becomes times

    • B.

      Increases K times

    • C.

      Remains unchanged

    • D.

      Decreases K times

    Correct Answer
    A. Becomes times
    Explanation
    When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance becomes K times. This is because the presence of a dielectric medium increases the electric field between the charges, which in turn increases the force of attraction between them. The constant K, also known as the dielectric constant, quantifies the extent to which the medium can increase the electric field. Therefore, the force of attraction between the charges becomes K times greater when air is replaced by a dielectric medium.

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  • 5. 

    Potential energy of two equal negative point charges 2µC each held 1m apart in air is:

    • A.

      2 J

    • B.

      2 eV

    • C.

      4 J

    • D.

      0.036 J

    Correct Answer
    D. 0.036 J
    Explanation
    The potential energy between two charges can be calculated using the formula U = k * (q1 * q2) / r, where U is the potential energy, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. In this case, the charges are negative and equal to -2 μC each, and the distance is 1m. Plugging these values into the formula, we get U = (8.99 * 10^9 N m²/C²) * (-2 μC * -2 μC) / 1m = 0.036 J.

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  • 6. 

    The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0 x 10-15 m is:

    • A.

      80 volt

    • B.

      8 x  volt

    • C.

      9 volt

    • D.

      9 x  volt

    Correct Answer
    B. 8 x  volt
  • 7. 

    Three different capacitors are connected in series. Then:

    • A.

      They will have equal charges

    • B.

      They will have same potential

    • C.

      Both 1 & 2

    • D.

      None of these

    Correct Answer
    A. They will have equal charges
    Explanation
    When capacitors are connected in series, the same amount of charge passes through each capacitor. This is because the capacitors are connected in a series circuit, so the current flowing through them is the same. Since the charge on a capacitor is directly proportional to the potential difference across it, and the potential difference is the same for capacitors connected in series, it follows that the charges on the capacitors will be equal. Therefore, option 1 is correct.

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  • 8. 

    A 2µF capacitor is charged to 100 volt and then its plates are connected by a conducting wire. The heat produced is:

    • A.

      0.001 J

    • B.

      0.01 J

    • C.

      0.1 J

    • D.

      1 J

    Correct Answer
    B. 0.01 J
    Explanation
    When the plates of the charged capacitor are connected by a conducting wire, the stored electrical energy in the capacitor is discharged. This discharge causes a flow of current through the wire, which leads to the generation of heat. The amount of heat produced can be calculated using the formula: Heat = 0.5 * C * V^2, where C is the capacitance and V is the voltage. In this case, the capacitance is given as 2µF and the voltage is 100 volts. Plugging these values into the formula gives us a heat of 0.01 J. Therefore, the correct answer is 0.01 J.

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  • 9. 

    A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 volt. The potential at the centre of the sphere is:

    • A.

      8 volt

    • B.

      800 volt

    • C.

      80 volt

    • D.

      Zero

    Correct Answer
    C. 80 volt
    Explanation
    The potential at the center of the sphere is 80 volts because the potential at any point inside a hollow metal sphere is constant and equal to the potential on its surface. Since the potential on the surface is given as 80 volts, the potential at the center will also be 80 volts.

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  • 10. 

    A parallel plate capacitor is made by stocking n equally spaced plates connected alternately. If the capacitance between any two plates is x, then the total capacitance is:

    • A.

      Nx

    • B.

      N/x

    • C.

      X

    • D.

      (n – 1) x

    Correct Answer
    D. (n – 1) x
    Explanation
    The total capacitance of a parallel plate capacitor made by stocking n equally spaced plates connected alternately can be found by multiplying the number of plates (n) minus 1, with the capacitance between any two plates (x). This is because each plate is connected to two neighboring plates, except for the first and last plates which are only connected to one neighboring plate. Therefore, the total capacitance is equal to (n - 1) times x.

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  • 11. 

    Three capacitors each of capacity 4 µF are to be connected in such a way that the effective capacitance is 6 µF. This can be done by:

    • A.

      Connecting all of them in series

    • B.

      Connecting all of them in parallel

    • C.

      Connecting two in series and one in parallel

    • D.

      Connecting two in parallel and one in series

    Correct Answer
    C. Connecting two in series and one in parallel
    Explanation
    Connecting two capacitors in series and one capacitor in parallel allows for the effective capacitance to be calculated using the formula 1/Ceff = 1/C1 + 1/C2 + 1/C3. In this case, if two capacitors are connected in series, their effective capacitance would be 2 µF. Then, by connecting the third capacitor in parallel to the series combination, the effective capacitance would be increased to 6 µF, as desired.

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  • 12. 

    The magnitude of electric field E in the annular region of a charged cylindrical capacitor.

    • A.

      Is the same throughout

    • B.

      Is higher near the outer cylinder than near the inner cylinder

    • C.

      Varies as 1/r, where r is the distance from the axis

    • D.

      Varies as 1/ , where r is the distance from the axis.

    Correct Answer
    C. Varies as 1/r, where r is the distance from the axis
    Explanation
    The correct answer is "Varies as 1/r, where r is the distance from the axis." In a charged cylindrical capacitor, the electric field magnitude decreases as the distance from the axis increases. This is because the electric field lines spread out as they move away from the axis, resulting in a decrease in field strength. The relationship between the electric field magnitude and the distance from the axis is inversely proportional, following the equation E ∝ 1/r.

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  • 13. 

    A 800 µF capacitor is charged at the steady rate of 50 µC/sec. How long will it take to raise its potential to 10 volt?

    • A.

      160 s

    • B.

      50 s

    • C.

      10 s

    • D.

      500 s

    Correct Answer
    A. 160 s
    Explanation
    The time it takes to raise the potential of a capacitor can be calculated using the formula: time = charge/capacitance. In this case, the charge is given as 50 µC/sec and the capacitance is 800 µF. By substituting these values into the formula, we get time = 50 µC/sec / 800 µF = 0.0625 sec. However, the question asks for the time in seconds, so we need to convert the time to seconds by multiplying it by 1000. Therefore, the correct answer is 0.0625 sec * 1000 = 62.5 s.

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  • 14. 

    If there are n capacitors in parallel connected to V volt source, then energy stored is equal to:

    • A.

      CV

    • B.

      C

    • C.
    • D.

      nC

    Correct Answer
    D. nC
    Explanation
    When capacitors are connected in parallel, the voltage across each capacitor is the same. The energy stored in a capacitor is given by the formula 1/2 * C * V^2, where C is the capacitance and V is the voltage. Since all the capacitors are connected in parallel to the same voltage source, the voltage V is constant for all capacitors. Therefore, the energy stored in each capacitor is equal to C * V. Since there are n capacitors in parallel, the total energy stored is equal to n * C * V, which can be simplified to nC.

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  • 15. 

    The equivalent capacity of the combination C1 = C2 = C3 shown in figure is:

    • A.

      2 C

    • B.

      C

    • C.

      C / 2

    • D.

      None of these

    Correct Answer
    A. 2 C
    Explanation
    The equivalent capacity of the combination C1 = C2 = C3 shown in the figure is 2C. This means that when these capacitors are connected in parallel, the total capacity is double the value of each individual capacitor.

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  • Current Version
  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Nov 21, 2013
    Quiz Created by
    Tanmay Shankar

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