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Physics M.C.Q (1 Marks)

Electrostatic Potential and Capacitance · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 274 questions.

Questions · Page 3 of 6

M.C.Q (1 Marks)

MCQ 1011 Mark
Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch S. If now the switch S is opened and the space between the plates is filled with a dielectric of relative permittivity εr​, then:

  • A
    The potential difference as well as charge on each capacitor goes up by a factor εr​.
  • B
    The potential difference as well as charge on each capacitor goes down by a factor εr.​
  • C
    The potential difference across A remains constant and the charge on B remains unchanged.
  • D
    The potential difference across B remains constant while the charge on A remains unchanged.
Answer
  1. The potential difference across A remains constant and the charge on B remains unchanged.

Explanation:

After switch S is opened, as the Capacitor A is connected across the battery, its potential difference is fixed at steady state (i.e., when capacitor is fully charged).

Capacitor B is isolated, so its charge gets fixed. But as we insert the dielectric, its capacitance changes, thus its potential difference also changes.

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MCQ 1031 Mark
Van de Graaff generator is used to:
  • A
    Store electrical energy.
  • B
    Build up high voltage of few million volts.
  • C
    Decelerate charged particle like electrons.
  • D
    Both (a) and (b).
Answer
  1. Build up high voltage of few million volts.
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MCQ 1041 Mark
A parallel plate capacitor is charged. If the plates are pulled apart:
  • A
    The capacitance increases.
  • B
    The potential difference increases.
  • C
    The total charge increases.
  • D
    The charge and potential difference remain the same.
Answer
  1. The potential difference increases.

Explanation:

Capacitance of a parallel plate capacitor.

$\text{C}=ε0\text{AdC}=ε0\text{Ad}$

A parallel plate capacitor is charged (battery is disconnected) then the plates are pulled apart, the capacitance decreases while the charge remains the same.

$\because$ Potentialdifference = ChargeCapacitance∵Potentialdifference = ChargeCapacitance.

$\therefore$ potential difference increases.

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MCQ 1051 Mark
A pendulum ( positively charged and hinged at some length above the plate) is swinging above a parallel plate (infinitely large and having negative charge), now consider the following statements, (consider gravity):
  • A
    Angular momentum about the hinge point of the ball will be max at lowest point.
  • B
    Electric potential energy will be max at highest point.
  • C
    Gravitational potential energy will be lowest at highest point.
  • D
    None of the above.
Answer
  1. Angular momentum about the hinge point of the ball will be max at lowest point.

Explanation:

Angular momentum about the hinge point of the ball will be max at point. mV will be max at closer point because law of conservation says that the swinging energy will be gathered when it will be at lowest point. The potential energy is maximum.

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MCQ 1061 Mark
A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge:
  • A
    Remains a constant because the electric field is uniform.
  • B
    Increases because the charge moves along the electric field.
  • C
    Decreases because the charge moves along the electric field.
  • D
    Decreases because the charge moves opposite to the electric field.
Answer
  1. Decreases because the charge moves along the electric field.

The positive charge will experience an electrostatic force whose direction will be along the direction of electric field.

In other words, positive charge will move from high electrostatic potential to low electrostatic potential.

Work will be done by electric filed on the charge and the electric potential energy of the charge will decrease.

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MCQ 1071 Mark
When air in a capacitor is replaced by a medium of dielectric constant K, the capacity:
  • A
    Decreases K times.
  • B
    Increases K times.
  • C
    Increases K2 times.
  • D
    Remains constant.
Answer
  1. Increases K times.
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MCQ 1091 Mark
 The electrostatic potential on the perpendicular bisector due to an electric dipole is _____________.
  • A
    Zero
  • B
    1
  • C
    Infinite
  • D
    Negative
Answer
  1. Zero

Explanation:

The electrostatic potential on the perpendicular bisector due to an electric dipole is zero.

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MCQ 1101 Mark
When moving electron comes closer to other stationary electron, then its kinetic energy and potential energy respectively _____ and _____.
  • A
    Increases, increases
  • B
    Increases, decreases
  • C
    Decreases, increases
  • D
    Decreases, decreases
Answer
  1. Decreases, increases

Explanation:

When electron comes closer to the other stationary electron, its kinetic energy decreases because of repulsion between them. As per conservation of energy, the potential energy increases.

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MCQ 1111 Mark
If the capacitors having capacitance C1 and C2 are connected in series then their resultant capacitance is given by:
  • A
    1/C = 1/C1 + 1/C2
  • B
    1/C = 1/C1 – 1/C2
  • C
    C = C1 + C2
  • D
    None
Answer
  1. 1/C = 1/C1 + 1/C2
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MCQ 1121 Mark
Work done by an external force in bringing a unit positive charge from infinity to a point is called as:
  • A
    Potential energy at that point.
  • B
    Electric field at that point.
  • C
    Electric potential at that point.
  • D
    None.
Answer
  1. Electric potential at that point.
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MCQ 1131 Mark
An insulator plate is passed between the plates of a capacitor. Then, current:

  • A
    First flows from A to B and then from B to A
  • B
    First flows from B to A and then from A to B
  • C
    Always flows from B to A
  • D
    Always flows from A to B
Answer
  1. First flows from B to A and then from A to B
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MCQ 1141 Mark
The plate current in a diode is zero. It is possible that:
  • A
    The plate voltage is zero.
  • B
    The plate voltage is slightly negative.
  • C
    The plate voltage is slightly positive.
  • D
    The temperature of the filament is low.
Answer
  1. The plate voltage is zero.
  2. The plate voltage is slightly negative.
  3. The plate voltage is slightly positive.
  4. The temperature of the filament is low.

Explanation:

The plate current varies directly with the plate voltage. Therefore, if the plate voltage is zero, the plate current is also zero. Due to the same reason, if the plate voltage is negative, the plate current will be zero. Now, if the plate is slightly positive, then it may be the reason that the plate voltage is not able to reduce the effect of space charge. Hence, the current may be zero. Now, as the temperature of the filament is low, it will not be able to emit electrons and the resulting plate current will be zero. Hence, all the options are possible.

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MCQ 1161 Mark
Two point charges 10C and -10C are placed at a certain distance. What is the electric potential of their midpoint?
  • A
    Some positive value.
  • B
    Some negative value.
  • C
    Zero.
  • D
    Depends on medium.
Answer
  1. Zero.

Explanation:

Electric potential is a scalar quantity and its value is solely dependent on the charge near it and the distance from that charge. In this case, the point is equidistant from the two point charges and the point charges have the same value but opposite nature. Therefore equal but opposite potentials are generated due to the charges and hence the net potential at midpoint becomes zero.

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MCQ 1171 Mark
Equipotential surfaces:
  • A
    Are closer in regions of large electric fields compared to regions of lower electric fields.
  • B
    Will be more crowded near sharp edges of a conductor.
  • C
    Will be more crowded near regions of large charge densities.
  • D
    Will always be equally spaced.
Answer
  1. Are closer in regions of large electric fields compared to regions of lower electric fields.
  2. Will be more crowded near sharp edges of a conductor.
  3. Will be more crowded near regions of large charge densities.

Equipotential surfaces are closer in regions of large electric fields because electric field intensity is inversely proportional to the separation between equipotential surfaces. 

As the electric field intensities is large near sharp edges of a charged conductor or near the regions of large charge densities. Therefore, numbers of equipotential surfaces are closer to such places or in other words they are more crowded.

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MCQ 1181 Mark
The electric potential at a point on the equatorial line of an electric dipole is:
  • A
    Directly proportional to distance.
  • B
    Inversely proportional to distance.
  • C
    Inversely proportional to square of the distance.
  • D
    None of these.
Answer
  1. None of these.
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MCQ 1191 Mark
Which of the following is / are true about the principle of Van de Graaff generator?
  • A
    The action of sharp points.
  • B
    The charge given to a hollow conductor is tranferred to outer surface and is distributed uniformly over it.
  • C
    It is used for accelerating uncharged particle.
  • D
    Both (a) and (b).
Answer
  1. Both (a) and (b).
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MCQ 1201 Mark
In a charged capacitor, the energy resides:
  • A
    In the positive charges.
  • B
    In both the positive and negative charges.
  • C
    In the field between the plates.
  • D
    Around the edges of the capacitor plates.
Answer
  1. In the field between the plates.
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MCQ 1211 Mark
The magnitude of the electric field is given by the change in magnitude of potential per unit displacement _____ to the equipotential surface at the point.
  • A
    Normal.
  • B
    Perpendicular.
  • C
    Parallel.
  • D
    Both a and b.
Answer
  1. Both a and b.
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MCQ 1221 Mark
When was Van de Graff generator invented and by whom?
  • A
    1944, Robin Van de Graff
  • B
    1932, Robert Van de Graff
  • C
    1933, Robin Van de Graff
  • D
    1933, Robert Van de Graff
Answer
  1. 1933, Robert Van de Graff

Explanation:

Van de Graff generator was invented by Robert Jemison Van de Graff on November 28, 1933. Robert Jemison invented the Van de Graff generator, which is a kind of high-voltage electrostatic generator that accelerates particles, while he was doing his PhD in Princeton University.

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MCQ 1231 Mark
The dipole moment per unit volume is called as:
  • A
    Dielectrics.
  • B
    Polarisation.
  • C
    Electric field.
  • D
    Electric dipole moment.
Answer
  1. Polarisation.
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MCQ 1241 Mark
A parallel plate air capacitor has capacity ′C′ farad, potential ′V′ volt and energy ′E′ joule. When the gap between the plates is completely filled with dielectric.
  • A
    Both V and E increase
  • B
    Both V and E decrease
  • C
    V decreases, E increases
  • D
    V increases, E decreases
Answer
  1. Both V and E decrease

Explanation:

A parallel- plate capacitor with a dielectric. The electric field is reduced between the plates because the dielectric material is polarized, producing an opposing field. When there is a dielectric, the potential is also reduced because potential is inversely proportional to dielectric

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MCQ 1251 Mark
1 Volt = _____________ style-type: lower- alpha;”>
  • A
    1 Coulomb
  • B
    1 Newton/ 1 Coulomb
  • C
    1 Joule/ 1 Coulomb
  • D
    1 Newton/ 1 meter
Answer
  1. 1 Joule/ 1 Coulomb
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MCQ 1261 Mark
From a point charge, there is a fixed point A. At A, there is an electric field of 500V/ m and potential difference of 3000V. Distance between point charge and A will be:
  • A
    6m.
  • B
    12m.
  • C
    16m.
  • D
    24m.
Answer
  1. 6m.
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MCQ 1281 Mark
A capacitor is a system of two conductors separated by _____.
  • A
    Conductors.
  • B
    Dielectrics.
  • C
    An insulators.
  • D
    None.
Answer
  1. An insulators.
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MCQ 1291 Mark
Capacity of a parallel plate condenser can be increased by:
  • A
    Increasing the distance between the plates.
  • B
    Increasing the thickness of the plates.
  • C
    Decreasing the thickness of the plates.
  • D
    Decreasing the distance between the plates.
Answer
  1. Decreasing the distance between the plates.
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MCQ 1301 Mark
Electrostatic potential is ______ throughout the volume of the conductor and has _____ value on its surface.
  • A
    Same, constant.
  • B
    Constant, same.
  • C
    Different, same.
  • D
    Constant, different.
Answer
  1. Constant, same.
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MCQ 1311 Mark
What is the electric field in the cavity of a hollow charged conductor?
  • A
    Positive
  • B
    Negative
  • C
    Zero
  • D
    Depends on the nature of the conductor
Answer
  1. Zero

Explanation:

By Gauss’s theorem, the charge enclosed by the gaussian surface is zero. Consequently, the electric field must be zero at every point inside the cavity. Then, the entire excess charge lies on its surface.

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MCQ 1321 Mark
1 electron volt = __________ J.
  • A
    1.6 × 10-19
  • B
    4.8 × 10-19
  • C
    1.6 × 10-10
  • D
    10
Answer
  1. 1.6 × 10-19

Explanation:

1 electron volt is the amount of work done if an electron is passed through a potential difference of 1V. Therefore the work done = 1V × charge of an electron = 1.602 × 10-19 J. But it is a small quantity and hence we use kilo electron volt and mega electron volt in practical.

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MCQ 1331 Mark
Following operations can be performed on a capacitor:
X - connect the capacitor to a battery of emf $\in.$
Y - disconnect the battery.
Z - reconnect the battery with polarity reversed.
W - insert a dielectric slab in the capacitor.
  • A
    In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.
  • B
    The charge appearing on the capacitor is greater after the action XWY than after the action XYW.
  • C
    The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
  • D
    The electric field in the capacitor after the action XW is the same as that after WX.
Answer
  1. The charge appearing on the capacitor is greater after the action XWY than after the action XYW.
  2. The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
  3. The electric field in the capacitor after the action XW is the same as that after WX.

Explanation:

Justification of option (b):

If the potential is held constant, that is, the battery remains attached to the circuit, then the charge on the capacitor increases by a factor of K on inserting a dielectric of a dielectric constant K between the plates of the capacitor. 

Mathematically,

q = Kq0 

Here, q0 and q are the charges without dielectric and with dielectric, respectively.

The amount of charge stored does not depend upon the polarity of the plates.

Thus, the charge appearing on the capacitor is greater after the action XWY than after the action XYZ.

Justification of option (c):

Since the battery is disconnected before inserting a dielectric, the amount of charge remains constant, that is, q = q​0, because after the battery is disconnected, the capacitor gets no source to store charge from. In other words, the capacitor is now an isolated system where the amount of charge is conserved and so is the energy U as $\frac{1}{2}\text{q}\in.$ Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor. So, the energy stored in the capacitor will also not change after the action XYW.

However, during the action WXY, the amount of charge that will get stored in the capacitor will get increased by a factor of K, as the battery is disconnected after inserting a dielectric between the plates of the capacitor and the energy stored will also get multiplied by a factor of K.

Thus, the electric energy stored in the capacitor is greater after the action WXY than after the action XYW.

Justification of option (d):

The electric field between the plates E depends on the potential across the capacitor and the distanced between the plates of the capacitor.

Mathematically, 

$\text{E}=\frac{\in}{\text{d}}$

In either case, that is, during actions XW and WX, the potential remains the same, that is, $\in.$ Thus, the electric field E remains the same.

Denial of option (a):

During the action XYZ, the battery has to do extra work equivalent to $\frac{1}{2}\text{CV}^2$ to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be $\frac{1}{2}\text{CV}^2+\frac{1}{2}\text{CV}^2.$

This extra work done will be dissipated as heat energy. Thus, thermal energy is developed. However, the stored electric energy remains unchanged, that is, $\frac{1}{2}\text{CV}^2.$

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MCQ 1341 Mark
The factor by which the capacitance increases from its vacuum value when the dielectric is inserted fully between the plates of a capacitor, is called as:
  • A
    Dielectric.
  • B
    Dielectric constant of the substance.
  • C
    Permittivity.
  • D
    Permiability.
Answer
  1. Dielectric constant of the substance.
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MCQ 1361 Mark
In parallel combination of capacitors, the effective capacitance:
  • A
    Decreases.
  • B
    Increases.
  • C
    Remains same.
  • D
    None.
Answer
  1. Increases.
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MCQ 1371 Mark
A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is_______?
  • A
    zero
  • B
    positive
  • C
    negative
  • D
    zero if the charge Q is at the center and nonzero otherwise.
Answer
  1. zero

Explanation:

The net displacement round one complete circle is 0.

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MCQ 1381 Mark
In case of a Van de Graaff generator, the breakdown field of air is:
  • A
    2 × 108V m–1
  • B
    3 × 106V m–1
  • C
    2 × 108V m–1
  • D
    3 × 104V m–1
Answer
  1. 3 × 106V m–1
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MCQ 1391 Mark
The work done by the external force in bringing the charge q from infinity to a point is called as:
  • A
    Electric field due to charge q at that point.
  • B
    Potential energy due to charge q at that point.
  • C
    Both a and b.
  • D
    None.
Answer
  1. Potential energy due to charge q at that point.
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MCQ 1401 Mark
Mark the correct options:
  • A
    A diode valve can be used as a rectifier.
  • B
    A triode valve can be used as a rectifier.
  • C
    A diode valve can be used as an amplifier.
  • D
    A triode valve can be used as an amplifier.
Answer

  1. A diode valve can be used as a rectifier.

  2. A triode valve can be used as a rectifier.

  1. A triode valve can be used as an amplifier.

Explanation:

A diode valve and a triode valve allow current to flow only in one direction. Since a rectifier is a device that converts alternating current (bi-directional) into direct current (uni-directional), a diode valve and a triode valve can be used as rectifiers. A triode valve can control its output in proportion to the input signal. That is, it can act as an amplifier, whereas a diode valve cannot control its output in proportion to the input signal. So, it cannot be use as an amplifier.

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MCQ 1411 Mark
A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is:
  • A
    $\frac{\text{CV}}{\in_0}$
  • B
    $\frac{\text{2CV}}{\in_0}$
  • C
    $\frac{\text{CV}}{2\in_0}$
  • D
    $\text{Zero}.$
Answer
  1. $\text{Zero}.$

Explanation:

Since the net charge enclosed by the Gaussian surface is zero, the total flux of the electric field through the closed Gaussian surface enclosing the capacitor is zero.

$\phi=\oint\text{E.ds}=\frac{\text{q}}{\in_0}=0$

Here,

$\phi=$ Electric flux

q = Total charge enclosed by the Gaussian surface.

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MCQ 1421 Mark
Equipotential surfaces of a single point charge are _____ surfaces centred at the charge.
  • A
    Concentric cylindrical.
  • B
    Concentric circular.
  • C
    Concentric spherical.
  • D
    None.
Answer
  1. Concentric spherical.
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MCQ 1431 Mark
Which of the following about potential at a point due to a given point charge is true?
The potential at a point P due to a given point charge.
  • A
    Is a function of distance from the point charge.
  • B
    Varies inversely as the square of distance from the point charge.
  • C
    Is a vector quantity.
  • D
    Is directly proportional to the square of distance from the point charge.
Answer
  1. Is a function of distance from the point charge.
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MCQ 1441 Mark
Coulomb’s force between two stationery charges is also:
  • A
    Non conservative force.
  • B
    Conservative force.
  • C
    Both a and b.
  • D
    None.
Answer
  1. Conservative force.
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MCQ 1451 Mark
Electric conduction takes place in a discharge tube due to the movement of:
  • A
    Positive ions.
  • B
    Negative ions.
  • C
    Electrons.
  • D
    Protons.
Answer
  1. Positive ions.
  2. Negative ions.
  3. Electrons.

Explanation:

Some ions are always present in gases due to cosmic rays and other factors. When we apply potential difference  across a discharge tube, the ions get accelerated due to the electric field. If the potential difference is large, then the ions get enough energy to ionise the molecules on collision. Thus, a large number of ions are produced and conduction starts. Generally, an electron gets detached from a molecule to make the molecule a positive ion. At low pressure, this electron moves through a large distance and gets attached to another molecule and forms a negative ion. Thus, electric conduction takes place in a discharge tube due to the movement of positive ions, negative ions and electrons.

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MCQ 1461 Mark
When the separation between two charges is increased the electric potential energy of the charges.
  • A
    Increases.
  • B
    Decreases.
  • C
    Remains the same.
  • D
    May increase or decrease.
Answer

  1. May increase or decrease.

Explanation:

When the separation between two charges is increased, the electric potential Energy of charge may incease or decrease.

If Both charge are like charge then electric potential energy of charge decreases.

$\text{U}=\frac{\text{k}\text{q}_1\text{q}_2}{\text{r}}$

If Both charge are unlike charge then electric potential energy of charge increases.

$\text{U}=\frac{-\text{kq}_1\text{q}_2}{\text{r}}$

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MCQ 1471 Mark
The positive terminal of 12V battery is connected to the ground. Then the negative terminal will be at:
  • A
    - 6V
  • B
    + 12V
  • C
    Zero
  • D
    - 12V
Answer
  1. - 12V
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MCQ 1481 Mark
Which of the following statement is not true?
  • A
    Electrostatic force is a conservative force.
  • B
    Potential at a point is the work done per unit charge in bringing a charge from infinity to that point in an electric field.
  • C
    Electrostatic force is non-conservative.
  • D
    Potential is the ratio of work to charge.
Answer
  1. Electrostatic force is non-conservative.

Explanation:

Work done by the electrostatic force is independent of the path followed by it, and it depends only on the initial and final positions. For example, work done in moving a unit positive charge in a closed loop of an electric field is zero. So,electrostatic force is a conservative force.

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MCQ 1491 Mark
 Dielectrics are _____________?
  • A
    Conducting substances.
  • B
    Non-conducting substances.
  • C
    Semi-conducting substances.
  • D
    None of the option.
Answer
  1. Non-conducting substances.

Explanation:

Dielectrics are non-conducting substances.

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