Question 511 Mark
A parallel plate condenser is immersed in an oil of dielectric constant 2. The field between the plates is:
Answer
View full question & answer→- Decreased, proportional to 1/ 2.
Questions · Page 2 of 6
M.C.Q (1 Marks)
Question 521 Mark
Capacitiors are used in electrical circuits where appliances need more:
Answer
View full question & answer→- Current.
Question 531 Mark
There are two metallic spheres of same radii, but one is solid, and the other is hollow, then:
Answer
View full question & answer→- They can be charged equally (maximum).
Question 541 Mark
1 nano farad =
Answer
View full question & answer→- 10-9.
Question 551 Mark
Three capacitors of capacitances $6\mu\text{F}$ each are available. The minimum and maximum capacitances, which may be obtained are:
Answer
The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:
$\frac{1}{\text{C}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$
$\Rightarrow\text{C}=2\mu\text{F}$
The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:
$\text{C}=6+6+6=18\mu\text{F}$
View full question & answer→- $2\mu\text{F},\ 18\mu\text{F}$
The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:
$\frac{1}{\text{C}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$
$\Rightarrow\text{C}=2\mu\text{F}$
The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:
$\text{C}=6+6+6=18\mu\text{F}$
Question 561 Mark
what is the potential difference between two points, if 2J of work must be done to move a 4 mC charge from one point to another is:
Answer
The total work done = energy transferred.
So, we might see the equation energy = voltage x charge, E = V × Q, written as,
work = voltage x charge, W = V × Q.
In this case, the charge is 4 mC, that is, 0.004 C and work done is 2J.
View full question & answer→- 500 V
The total work done = energy transferred.
So, we might see the equation energy = voltage x charge, E = V × Q, written as,
work = voltage x charge, W = V × Q.
In this case, the charge is 4 mC, that is, 0.004 C and work done is 2J.
Question 571 Mark
Which of the following is an example of a molecule whose centre of mass of positive and negative charges does not coincide each other?
Answer
View full question & answer→$NH_3$ is a molecule in which the centre of mass of positive and negative charges does not collide with each other and is called a polar molecule. They have a permanent dipole moment. They have unsymmetrical shapes.
Question 581 Mark
An example of an equipotential surface in earth is:
Answer
The potential is seen to be a constant on a sphere at all points.
View full question & answer→- A spherical surface at a distance of 1km from the surface of the earth with its centre at centre of earth.
The potential is seen to be a constant on a sphere at all points.
Question 591 Mark
For any charge configuration, equipotential surface through a point is _____ to electric field at that point.
Answer
View full question & answer→- Both a and b.
Question 601 Mark
Work done to bring a unit positive charge from infinity to a point in an electric field is known as _______?
Answer
Electric potential is defined as the amount of work done to bring a unit positive charge from an infinite distance to a particular point of an electric field. The total energy of that point charge means the sum of kinetic energy and potential energy which is not the same as the potential energy if the particle is in motion.
View full question & answer→- Electric potential.
Electric potential is defined as the amount of work done to bring a unit positive charge from an infinite distance to a particular point of an electric field. The total energy of that point charge means the sum of kinetic energy and potential energy which is not the same as the potential energy if the particle is in motion.
Question 611 Mark
What type of surface is the surface of a conductor?
Answer
Electric field at any point is equal to the negative of the potential gradient. But inside a conductor, the electric field is zero. Hence, the electric potential is constant throughout the volume of a conductor and has the same value on its surface. Thus the surface of a conductor is equipotential.
View full question & answer→- Equipotential
Electric field at any point is equal to the negative of the potential gradient. But inside a conductor, the electric field is zero. Hence, the electric potential is constant throughout the volume of a conductor and has the same value on its surface. Thus the surface of a conductor is equipotential.
Question 621 Mark
A slab $X$ is placed between the two parallel isolated charged plates as shown in the figure. If $E_p$ and $E_q$ denotes the intensity of electric field at$ P$ and $Q,$ then:
Answer
View full question & answer→There is no effect of the metal on the external electric field $($ thus $A$ is incorrect$)$ while the dielectric reduces the net electric field outside. The dielectric produces an electric field inside it due to the induced charges which is opposite to the external field. Thus $EQ$ is reduced and $B$ and $C$ are also incorrect.$($Note that $E_P$ is not the external field$).$ Now as the electric field inside the conductor is zero, the field $E_{p}$ is zero if $X$ is a metallic.
Question 631 Mark
What is the nature of equipotential surfaces in case of a positive point charge?
Answer
We know that electric field lines cross the equipotential surfaces perpendicularly. Electric field lines are generated radially from a positive point charge. Therefore, for holding both the conditions, the equipotential surfaces must be spherical.
View full question & answer→- Spherical
We know that electric field lines cross the equipotential surfaces perpendicularly. Electric field lines are generated radially from a positive point charge. Therefore, for holding both the conditions, the equipotential surfaces must be spherical.
Question 641 Mark
Potential due to charge q at its own location is:
Answer
View full question & answer→- Infinite.
Question 651 Mark
If the capacitors having capacitance $C_1$ and $C_2$ are connected in parallel then their effective capacitance is given by:
Answer
View full question & answer→$C = C_1 + C_2$
Question 661 Mark
Potential energy is the characteristic of the:
Answer
View full question & answer→- The present state of the configuration.
Question 671 Mark
The net charge on a condenser is:
Answer
Every condenser is made with two plates. The charge on one plate is +Q and other is −Q.
Thus total charge of condenser is Qt = +Q - Q = 0
View full question & answer→- zero
Every condenser is made with two plates. The charge on one plate is +Q and other is −Q.
Thus total charge of condenser is Qt = +Q - Q = 0
Question 681 Mark
1 volt = ?
Answer
Using: $\text{V}=\frac{\text{W}}{\text{q}}$
⟹ 1 volt is equal to 1 joule per coulomb
View full question & answer→- 1 joule per coulomb.
Using: $\text{V}=\frac{\text{W}}{\text{q}}$
⟹ 1 volt is equal to 1 joule per coulomb
Question 691 Mark
A charge q is at the center of the circle ABCDE. Which among the following is true if the charge is taken from A to B, C, D, and E?
Answer
As the charge q is situated at the center of the circle ABCDE, therefore the circle is an equipotential surface. That means all the points on the circle i.e. A, B, C, D, and E have the same potential. Therefore, work done to bring the charge from A to any point on the circle is zero always.
View full question & answer→- Work done along all the paths are zero.
As the charge q is situated at the center of the circle ABCDE, therefore the circle is an equipotential surface. That means all the points on the circle i.e. A, B, C, D, and E have the same potential. Therefore, work done to bring the charge from A to any point on the circle is zero always.
Question 701 Mark
1 micro farad =
Answer
View full question & answer→- 10-6
Question 711 Mark
What is the amount of work done to bring a charge of $4 \times 10^{-3}C$ charge from infinity to a point whose electric potential is $2 \times 10^2V?$
Answer
View full question & answer→Work done $=$ potential \times charge by definition. We know that the potential of a point is the amount of work done to bring a unit charge from infinity to a certain point. Therefore, work done $W = q \times V = 4 \times 10^{-3 }\times 200J = 0.8J.$ The work done is positive in this case.
Question 721 Mark
The process in which a region is made free from any electric field is known as _____________.
Answer
Electrostatic shielding is a phenomenon seen when a Faraday cage is used to block the effects of an electric field.
View full question & answer→- Electrostatic shielding
Electrostatic shielding is a phenomenon seen when a Faraday cage is used to block the effects of an electric field.
Question 731 Mark
What is the electric potential at the perpendicular bisector of an electric dipole?
Answer
Any point on the perpendicular bisector is equidistant from both the charges of the dipole. Therefore, the electric potential at that point is equal and opposite due to the two different charges. Therefore, the total electric potential at that point is zero.
View full question & answer→- Zero
Any point on the perpendicular bisector is equidistant from both the charges of the dipole. Therefore, the electric potential at that point is equal and opposite due to the two different charges. Therefore, the total electric potential at that point is zero.
Question 741 Mark
If a charged body is moved in an electric field against the Coulomb force, then ________?
Answer
To move a body against some force, work is to be done on the body. In this case, an external force is to be applied on the body to move it i.e. an external work is to be done. As we are moving the body against the Coulomb’s force, hence no work is done on the body by the electric field.
View full question & answer→- Work is done on the body by an external agent.
To move a body against some force, work is to be done on the body. In this case, an external force is to be applied on the body to move it i.e. an external work is to be done. As we are moving the body against the Coulomb’s force, hence no work is done on the body by the electric field.
Question 751 Mark
A capacitor of 4μ F is connected as shown in the circuit (Fig.). The internal resistance of the battery is 0.5Ω. The amount of charge on the capacitor plates will be:
Answer
The potential difference across 10Ω resistance will be zero. It will act like a plain wire.
Current flowing through 2Ω resistance is given by
$\text{I}=\frac{\text{V}}{(\text{R}+\text{r})}=\frac{2.5}{(2+0.5)}=1\text{A}$
Potential difference across 2Ω resistance V = IR = 1 × 2 = 2V
Here, capacitor is connected in parallel with 2 Ω resistance, so it will also have 2V potential difference across it.
The charge on capacitor, q = CV = (2 mF) × 2V = 8 mC.
View full question & answer→- 8μ C.
The potential difference across 10Ω resistance will be zero. It will act like a plain wire.
Current flowing through 2Ω resistance is given by
$\text{I}=\frac{\text{V}}{(\text{R}+\text{r})}=\frac{2.5}{(2+0.5)}=1\text{A}$
Potential difference across 2Ω resistance V = IR = 1 × 2 = 2V
Here, capacitor is connected in parallel with 2 Ω resistance, so it will also have 2V potential difference across it.
The charge on capacitor, q = CV = (2 mF) × 2V = 8 mC.
Question 761 Mark
A parallel plate condenser is connected to a battery of emf 4 volt. If a plate of dielectric constant 8 is inserted into it, the potential difference on the condenser will be:
Answer
Total charged is conserved in the condenser. Here the potential difference on the condenser is equal to the emf of battery. When dielectric is inserted between the plates the charge will maintain the constant potential in the capacitor. Thus, the potential difference on the condenser is 4V
View full question & answer→- 4V
Total charged is conserved in the condenser. Here the potential difference on the condenser is equal to the emf of battery. When dielectric is inserted between the plates the charge will maintain the constant potential in the capacitor. Thus, the potential difference on the condenser is 4V
Question 771 Mark
When a dielectric slab is introduced between the two plates of condenser then its capacity ___________.
Answer
As the dielectric slab is introduced there is some charge distribution in the slab and because of this the electric field between the two plates is decreased, due to which the capacitor can hold more charge. Thus, the capacity to hold charge of the capacitor is increased.
View full question & answer→- Increases
As the dielectric slab is introduced there is some charge distribution in the slab and because of this the electric field between the two plates is decreased, due to which the capacitor can hold more charge. Thus, the capacity to hold charge of the capacitor is increased.
Question 781 Mark
Work done in moving an object through an equipotential surface is:
Answer
Work done is given difference in potentials. In an equipotential surface, all points will have same potential. Thus work done is zero
View full question & answer→- Zero
Work done is given difference in potentials. In an equipotential surface, all points will have same potential. Thus work done is zero
Question 791 Mark
. The electric potential inside a conducting sphere _____________?
Answer
View full question & answer→- remains constant from centre to the surface.
Question 801 Mark
Which of the following is correct statement:
Answer
There is no potential gradient along any direction parallel to the surface, and no electric field is parallel with the surface, This means electric field are always at right angle to the equipotential surface.
View full question & answer→- Equipotential lines are always perpendicular to the electric field.
There is no potential gradient along any direction parallel to the surface, and no electric field is parallel with the surface, This means electric field are always at right angle to the equipotential surface.
Question 811 Mark
The molecules in which centres of positive and negative charges coincide are called as:
Answer
View full question & answer→- Non polar molecules.
Question 821 Mark
Point A is at a lower electrical potential than point B. An electron between them on the line joining them will:
Answer
Given that point A is at lower electric potential than point B. The electron between them on line joining will move.
We have to find where this electron moves.
Since we know that electric currents move from a higher potential or a lower potential. Also, electrons move in the direction opposite to electric current. So the electron on the line joining two points A and B will move from lower to higher potential i.e, it will move towards B.
View full question & answer→- Move towards B
Given that point A is at lower electric potential than point B. The electron between them on line joining will move.
We have to find where this electron moves.
Since we know that electric currents move from a higher potential or a lower potential. Also, electrons move in the direction opposite to electric current. So the electron on the line joining two points A and B will move from lower to higher potential i.e, it will move towards B.
Question 831 Mark
How is the electric field at the surface of a charged conductor related to the surface charge density?
Answer
The electric field at the surface of a charged conductor is proportional to the surface charge density. The electric field is zero inside the conductor and just outside, it is normal to the surface. The contribution to the total flux comes only from its outer cross-section.
View full question & answer→- Proportional to each other.
The electric field at the surface of a charged conductor is proportional to the surface charge density. The electric field is zero inside the conductor and just outside, it is normal to the surface. The contribution to the total flux comes only from its outer cross-section.
Question 841 Mark
The work done in placing a charge of $8 \times 10^{-18}$ coulomb on a condenser of capacity $100$ micro$-$farad is:
Answer
View full question & answer→$32 \times 10^{-32}$ joule.
Question 851 Mark
The work done to move a charge along an equipotential from $A$ to $B:$
Answer
View full question & answer→- Must be defined as $-\int\limits_\text{A}^\text{B}\text{E}.\text{dl}$.
- Is zero.
Work done in moving a charge particle from point $A$ to point $B$ is given by $W_{AB} = q(V_B - V_A)$ and the line integral of electrical field from point $A$ to $B$ gives potential difference i.e., $\text{V}_\text{B}-\text{V}_\text{A}=-\int\limits_\text{A}^\text{B}\text{E}.\text{dl}$
Now, work done to move a charge aling an equipotential from $A$ to $B$ is zero when potentials at point $A$ and $B$ are some.
Now, work done to move a charge aling an equipotential from $A$ to $B$ is zero when potentials at point $A$ and $B$ are some.
Question 861 Mark
The work done in moving a unit positive test charge over a closed path in an electric field is _____________.
Answer
We say electrostatic forces are conservative in nature since the work done in moving a unit positive test charge over a closed path in an electric field is zero.
View full question & answer→- Zero
We say electrostatic forces are conservative in nature since the work done in moving a unit positive test charge over a closed path in an electric field is zero.
Question 871 Mark
Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately.
Answer
View full question & answer→- Spheres.
Question 881 Mark
Non polar molecules has:
Answer
View full question & answer→- No permanent dipole moment.
Question 891 Mark
$1V$ equals:
Answer
View full question & answer→$1$ volt is equal to $1$ joule of electric potential energy per $($divided by$) 1$ coulomb of charge.
$V= J/ C =$ joule/ coulomb
$V= J/ C =$ joule/ coulomb
Question 901 Mark
How does the capacitance change with the effect of the dielectric when the battery remains connected across the capacitor?
Answer
View full question & answer→When a dielectric is introduced, and the battery remains connected across the capacitor, the capacitance increases from $C_0$ to $C.$
$C = kC_0.$
$C = kC_0.$
Question 911 Mark
If a conductor has a potential V ≠ 0 and there are no charges anywhere else outside, then:
Answer
Hence option (a) is correct. The charge resides on the outer surface of a closed charged conductor. Hence there cannot be any charge in the body of the conductor.
Hence option (b) is correct.
View full question & answer→- There must be charges on the surface or inside itself.
- There cannot be any charge in the body of the conducor.
Hence option (a) is correct. The charge resides on the outer surface of a closed charged conductor. Hence there cannot be any charge in the body of the conductor.
Hence option (b) is correct.
Question 921 Mark
The capacity of a parallel plate air condenser is 2 μF. If a dielectric of dielectric constant 4 is introduced between the plates, its new capacity is:
Answer
We know $\text{C}=\frac{\epsilon_0\text{A}}{\text{d}}$
When dielectric is added:
$\text{C'}=\frac{\text{K}\epsilon_0\text{A}}{\text{d}}=4\times2\mu\text{F}=8\mu\text{F}$
View full question & answer→- 8 μF
We know $\text{C}=\frac{\epsilon_0\text{A}}{\text{d}}$
When dielectric is added:
$\text{C'}=\frac{\text{K}\epsilon_0\text{A}}{\text{d}}=4\times2\mu\text{F}=8\mu\text{F}$
Question 931 Mark
What is the S.I. unit of electric potential?
Answer
An electric potential (also called the electric field potential or the electrostatic potential) is the amount of electric potential energy that a unitary point electric charge would have if located at any point of space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity to that point.
This value can be calculated in either a static (time-invariant) or a dynamic (varying with time) electric field at a specific time in units of joules per coulomb, or volts (V). The electric potential at infinity is assumed to be zero.
View full question & answer→- Volt
An electric potential (also called the electric field potential or the electrostatic potential) is the amount of electric potential energy that a unitary point electric charge would have if located at any point of space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity to that point.
This value can be calculated in either a static (time-invariant) or a dynamic (varying with time) electric field at a specific time in units of joules per coulomb, or volts (V). The electric potential at infinity is assumed to be zero.
Question 941 Mark
A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically Neutral.
Answer
View full question & answer→- A potential difference appears between the two cylinders when a charge density is given to the inner cylinder.
Question 951 Mark
An equipotential surface is that surface:
Answer
View full question & answer→- On which each and every point has the same potential.
Question 961 Mark
If the plates of a parallel plate charged capacitor are not parallel, the interface charge density is:
Answer
View full question & answer→- is non-uniform.
Question 971 Mark
The anode of a thermionic diode is connected to the negative terminal of a battery and the cathode to its positive terminal:
Answer
If the anode is given a negative potential relative to the cathode, the electrons are pushed back to the cathode. Hence, no current will flow through the diode.
View full question & answer→- No appreciable current will pass through the diode.
If the anode is given a negative potential relative to the cathode, the electrons are pushed back to the cathode. Hence, no current will flow through the diode.
Question 981 Mark
The capacitance of the capacitor is directly proportional to the:
Answer
View full question & answer→- Q.
Question 991 Mark
The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?
Answer
Because the charge always remains conserved in an isolated system, it will remain the same.
Now,
$\text{V}=\frac{\text{Qd}}{\epsilon_0\text{A}}$
Here, Q, A and d are the charge, area and distance between the plates, respectively.
Thus, as d increases, V increases.
Energy is given by:
$\text{E}=\frac{\text{qV}}{2}$
So, it will also increase.
Energy density u, that is, energy stored per unit volume in the electric field is given by:
$\text{u}=\frac{1}{2}\epsilon_0\text{E}^2$
So, u will remain constant with increase in distance between the plates.
View full question & answer→- Potential difference across the capacitor.
- Energy of the capacitor.
Because the charge always remains conserved in an isolated system, it will remain the same.
Now,
$\text{V}=\frac{\text{Qd}}{\epsilon_0\text{A}}$
Here, Q, A and d are the charge, area and distance between the plates, respectively.
Thus, as d increases, V increases.
Energy is given by:
$\text{E}=\frac{\text{qV}}{2}$
So, it will also increase.
Energy density u, that is, energy stored per unit volume in the electric field is given by:
$\text{u}=\frac{1}{2}\epsilon_0\text{E}^2$
So, u will remain constant with increase in distance between the plates.
Question 1001 Mark
The capacitance of the capacitor is given by:
Answer
View full question & answer→- C = Q/V.