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Physics M.C.Q (1 Marks)

Electric Field and Potential · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 17 questions.

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M.C.Q (1 Marks)

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17 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
If a body is charged by rubbing it, its weight
  1. Remains precisely constant.
  2. Increases slightly.
  3. Decreases slightly.
  4. May increase slightly or may decrease slightly.
Answer
  1. May increase slightly or may decrease slightly.

Explanation:

If a body is charged by rubbing it, its weight may increase slightly or may decrease slightly.

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Question 21 Mark
The electric potential decreases uniformly from 120V to 80V as one moves on the x-axis from x = -1cm to x = + 1cm. The electric field at the origin.
  1. Must be equal to 20Vcm-1
  2. May be equal to 20Vcm-1
  3. May be greater than 20Vcm-1
  4. May be lees than 20Vcm-1
Answer
  1. May be equal to 20Vcm-1
  2. May be greater than 20Vcm-1

Explanation:

$\triangle\text{V}=-\text{E}.\text{dr}$

$(\text{V}_\text{f}-\text{V}_\text{i})=-\text{E}.\text{dr}=\text{E}_\text{x}(\text{B}-\text{A})$

$(80-120)=-\text{E}_\text{x}.(2)$

$\text{E}_\text{x}=\frac{40}{2}=20\text{v/m}$

If electric field lines lies in 'x ' direction than it may be equal to 20v/cm.

If Electric field lines liesin 'x - y' direction than it may be greater than 20v/cm.

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Question 31 Mark
The electric field and the electric potential at a point are E and V respectively.
  1. If E = 0, V must be zero.
  2. If V = 0, E must be zero.
  3. If $\text{E}\neq0,$ V cannot be zero.
  4. If $\text{V}\neq0,$ E cannot be zero.
Answer
  1. If V = 0, E must be zero.

Explantion:

Electric field is negative gradient of electric potential.

$\text{E}=-\text{grad}(\text{V})$

$\text{E}=-\frac{\text{dV}}{\text{dr}}$

If $\text{E}=0$

$-\frac{\text{dV}}{\text{dr}}=0$

This implies

V = constant.

A constant can be zero or a quantifiable number.

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Question 41 Mark
Figure, shows some of the electric field lines corresponding to an electric field. The figure suggests that,

  1. EA > E8 > Ee
  2. EA = EB = Ee
  3. EA = Ec > EB
  4. EA = Ec < EB.
Answer

  1. EA = Ec > EB

Explanation:

Higher separation, Lower electric field. Because Electric field inversly proportional the square of separation.

EA = Ec > EB

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Question 51 Mark
When the separation between two charges is increased the electric potential energy of the charges.
  1. Increases.
  2. Decreases.
  3. Remains the same.
  4. 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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Question 61 Mark
Which of the following quantities do not depend on the choice of zero potential or zero potential energy?
  1. Potential at a point.
  2. Potential difference between two points.
  3. Potential energy of a two-charge system.
  4. Change in potential energy of a two-charge system.
Answer
  1. Potential energy of a two-charge system.
  2. Change in potential energy of a two-charge system.

Explanation:

$\text{V}_\text{p}=\frac{\text{KQ}}{\text{r}}-\frac{\text{KQ}}{\text{r}}=0$

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Question 71 Mark
Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential.
  1. Continuously increases.
  2. Continuously decreases.
  3. Increases then decreases.
  4. Decreases then increases.
Answer
  1. Continuously increases.

Explanation:

$\text{V}=\frac{\text{KQ}}{\text{r}}$

V → Electric Potential

$\text{V}_\text{p}=\frac{\text{KQ}}{\text{x}}+\frac{\text{KQ}}{\text{r}-\text{x}}=\frac{\text{KQ}\text{r}}{\text{x}(\text{r}-\text{x})}$

$\frac{\text{dvp}}{\text{dx}}=\frac{-\text{kQr}(\text{r}-2\text{x})}{\big(\text{r}(\text{r}-\text{x})\big)^2}=0$

$\text{r}=2\text{x},\ \text{x}=\frac{\text{r}}{2}$

$\text{V}_\text{Pmin}=\frac{\text{KQ}}{\frac{\text{r}}{2}}+\frac{\text{KQ}}{\frac{\text{r}}{2}}=\frac{4\text{KQ}}{\text{r}}$ at $\Big(\text{x}=\frac{\text{r}}{2}\Big)$

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Question 81 Mark
Mark out the correct options.
  1. The total charge of the universe is constant.
  2. The total positive charge of the universe is constant.
  3. The total negative charge of the universe is constant.
  4. The total number of charged particles in the universe is constant.
Answer

  1. The total charge of the universe is constant.

Explanation:

The total charge (Positive + negative) oif the universe is constant.

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Question 91 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 roatating charge in one complete revolution is:
  1. Zero.
  2. Positive.
  3. Negative.
  4. Zero if the charge Q is at the centre and nonzero otherwise.
Answer

  1. Zero.

Explanation:

In a circle Electric field due tot 'Q' is always perpendicular to the displacement oifthe charge 'q'.

Workdone $=\vec{\text{F}}.\vec{\text{d}}$

$=\big(\text{q}\vec{\text{E}}\big).\vec{\text{d}}=\text{q}\big(\vec{\text{E}}.\vec{\text{d}}\big)=0$

$\therefore\vec{\text{E}}\ \bot\ \vec{\text{d}}$

Workdoen by the Electric field on the rotating charge in one complete revolution is zero.

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Question 101 Mark
The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero,
  1. It is uniform in the region.
  2. It is proportional to r.
  3. It is proportional to r2.
  4. It increases as one goes away from the origin.
Answer

  1. It is proportional to r2.

Explanation:

$\text{DV}=-\text{E}.\text{dr}$

Given $(\text{E}\propto\text{r})$

$(\text{V}-0)=-\text{E}.\text{dr}$ 

$þ\text{E}\propto\text{r}^2$

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Question 111 Mark
A point charge is brought in an electric field. The electric field at a nearby point.
  1. Will increase if the charge is positive.
  2. Will decrease if the charge is negative.
  3. May increase if the charge is positive.
  4. May decrease if the charge is negative.
Answer
  1. May increase if the charge is positive.
  2. May decrease if the charge is negative.

Explanation:

The electric field at a near by point may increase if the charge is positive or may decreases if the charge is negative.

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Question 121 Mark
If a positive charge is shifted from a low-potential region to a high-potential region, the electric potential energy.
  1. Increases.
  2. Decreases.
  3. Remains the same.
  4. May increase or decrease.
Answer

  1. Increases.

Explanation:

Electric Potential Energy $=\text{q}\text{Dv}$

$=\text{q}(\text{v}_\text{f}-\text{v}_\text{i})$

If positive charge is shifted from a Low potential region to a High-Potential region, then electric Potential Energy increases.

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Question 131 Mark
A proton and an electron are placed in a uniform electric field.
  1. The electric forces acting on them will be equal.
  2. The magnitudes of the forces will be equal.
  3. Their accelerations will be equal.
  4. The magnitudes of their accelerations will be equal.
Answer
  1. The magnitudes of the forces will be equal.

Explanation:

Proton contain positive charge = 1.6 × 10-19 = e

Electron contain negative charge = -1.6 × 10-19 = -e

Force inoniforma electric field is = qE

$\therefore\ $The Magniludes of Electric force is equal But direction of electric force is opposite to each other.

mass of proton = 1.67 × 10-27Kg.

mass of electron = 9.1 × 10-31Kg.

So that Magnitudes of their acceleration will be unequal.

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Question 141 Mark
The electric field at the origin is along the positive x-axis. A small circle is drawn with the centre at the origin cutting the axes at points A, B, C and D having coordinates (a, 0), (0, a), (-a, 0), (0, -a) respectively. Out of the points on the periphery of the circle, the potential is minimum at.
  1. A
  2. B
  3. C
  4. D
Answer
  1. A

Explanation:

$\text{E}=\text{E}_0\hat{\text{i}}$ (Given)

(Potential at A):-

$\text{V}_\text{A}=\big(\text{E}_0\hat{\text{i}}\times\text{a}\hat{\text{i}}\big)=\text{E}_09$

(Potential atB):-

$\text{V}_\text{B}=\big(\text{E}_0\hat{\text{i}}\times\text{a}\hat{\text{i}}\big)=0$

(Potential at C):-

$\text{V}_\text{C}=\big(\text{E}_0\hat{\text{i}}\times(-\text{a})\text{i}\big)=\text{E}_0\text{a}$

(Potential atD):-

$\text{V}_\text{D}=(\text{E}_0\hat{\text{i}}\times(-\text{a})\hat{\text{j}})=0$

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Question 151 Mark
An electric dipole is placed in a uniform electric field. The net electric force on the dipole.
  1. Is always zero.
  2. Depends on the orientation of the dipole.
  3. Can never be zero.
  4. Depends on the strength of the dipole.
Answer

  1. Is always zero.

Explanation:

Net Electric force = FA + FB

= -qE + qE

= 0

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Question 161 Mark
An electric dipole is placed in an electric field generated by a point charge.
  1. The net electric force on the dipole must be zero.
  2. The net electric force on the dipole may be zero.
  3. The torque on the dipole due to the field must be zero.
  4. The torque on the dipole due to the field may be zero.
Answer
  1. The torque on the dipole due to the field may be zero.

Explanation:

In the uniform Electric field the net electric force on the dipole is alwas zero.

In uniform Electric field the torque on the dipole due to field may be zero.

T = 0

Here t Þ Torque

$\text{T}=2\text{qEl}\sin\text{q }\otimes\neq0$

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Question 171 Mark
Consider the situation of figure. The work done in taking a point charge from P to A is WA, from P to B is WB and from P to C is WC.

  1. WA < WB < WC
  2. WA > WB > WC
  3. WA = WB = WC
  4. None of these.

Answer

  1. WA = WB = WC

Explanation:

Electric Potential at A 'due to q' þ $\text{V}_\text{A}=\frac{\text{Kq}}{\text{r}}$

Electric Potetial at B 'due to q' þ $\text{V}_\text{B}=\frac{\text{Kq}}{\text{r}}$

& Electric potential at c 'due to q' þ $\text{V}_\text{C}=\frac{\text{Kq}}{\text{r}}$

Work done $=-\text{D}_\text{u}=-\text{q}\text{DV}$ {Let at 'P', Vp = 0}

Here $\text{V}_\text{A}=\text{V}_\text{B}=\text{V}_\text{C}$

The work done is taking a point charge from P to A, B & C is same.

So, $\text{W}_\text{A}=\text{W}_\text{B}=\text{W}_\text{C}$

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M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip