The electric potential at a point (x, y, z) is given by V = -x 2 y - xz 3 + 4. The electric field E → at that point is:

E → = (2xy + z 3 ) i ^ + x 2 j ^ + 3xz 2 k ^

E → = 2xy i ^ + (x 2 +y 2 ) j ^ +(3xz-y 2 ) k ^

E → = z 3 i ^ + xyz j ^ + z 2 k ^

E → = (2xy- z 3 ) i ^ + xy 2 j ^ + 3z 2 x k ^

Physics - Electrostatic Potential Capacitance - Quiz-10

In a region the potential is represented by V(x, y, z) = 6x – 8xy –8y + 6yz, where V is in volts and x, y, z, are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1,1) is :

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$6 \sqrt{5} N$ $6 \sqrt{5} N$
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30 N
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24 N
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$4 \sqrt{35} N$ $4 \sqrt{35} N$

A, B and C are three points in a uniform electric field. The electric potential is:

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maximum at B
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maximum at C
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same at all the three points A, B and C
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maximum at A

An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle $θ$ $θ$ with the direction of the field. Assuming that the potential energy of the dipole to be zero when $θ = 90 °$ $θ = 90 °$ , the torque and the potential energy of the dipole will respectively be

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$p E \sin θ, - p E \cos θ$ $p E \sin θ, - p E \cos θ$
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$p E \sin θ, - 2 p E \cos θ$ $p E \sin θ, - 2 p E \cos θ$
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$p E \sin θ, 2 p E \cos θ$ $p E \sin θ, 2 p E \cos θ$
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$p E \cos θ, - p E \sin θ$ $p E \cos θ, - p E \sin θ$

Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be:

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$\frac{C}{3}, \frac{V}{3}$ $\frac{C}{3}, \frac{V}{3}$
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$3 C, \frac{V}{3}$ $3 C, \frac{V}{3}$
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$\frac{C}{3}, 3 V$ $\frac{C}{3}, 3 V$
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3C, 3V

The electric potential at a point (x, y, z) is given by V = -x²y - xz³ + 4.
The electric field

\vec{E}

$\vec{E}$ at that point is:

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$\vec{E}$ $\vec{E}$ = (2xy + z³) $\hat{i}$ $\hat{i}$ + x² $\hat{j}$ $\hat{j}$ + 3xz² $\hat{k}$ $\hat{k}$
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$\vec{E}$ $\vec{E}$ = 2xy $\hat{i}$ $\hat{i}$ + (x² +y²) $\hat{j}$ $\hat{j}$ +(3xz-y²) $\hat{k}$ $\hat{k}$
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$\vec{E}$ $\vec{E}$ = z³ $\hat{i}$ $\hat{i}$ + xyz $\hat{j}$ $\hat{j}$ + z² $\hat{k}$ $\hat{k}$
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$\vec{E}$ $\vec{E}$ = (2xy- z³) $\hat{i}$ $\hat{i}$ + xy² $\hat{j}$ $\hat{j}$ + 3z²x $\hat{k}$ $\hat{k}$

The electric potential at a point in free space due to a charge Q coulomb is Q x 10¹¹ V. The electric field at that point is:

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$4 {πε}_{0} Q \times 10^{22} V / m$ $4 {πε}_{0} Q \times 10^{22} V / m$
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$12 {πε}_{0} Q \times 10^{20} V / m$ $12 {πε}_{0} Q \times 10^{20} V / m$
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$4 {πε}_{0} Q \times 10^{20} V / m$ $4 {πε}_{0} Q \times 10^{20} V / m$
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$12 {πε}_{0} Q \times 10^{22} V / m$ $12 {πε}_{0} Q \times 10^{22} V / m$

Two condensers, one of capacity C and the other of capacity C/2 are connected to a V volt battery, as shown.

The energy stored in the capacitors when both the condensers are fully charged:

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$2 C V^{2}$ $2 C V^{2}$
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$\frac{1}{4} C V^{2}$ $\frac{1}{4} C V^{2}$
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$\frac{3}{4} C V^{2}$ $\frac{3}{4} C V^{2}$
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$\frac{1}{2} C V^{2}$ $\frac{1}{2} C V^{2}$

Charges +q and –q are placed at points A and B, respectively; which are at a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is :

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$\frac{q Q}{4 {πε}_{0} L}$ $\frac{q Q}{4 {πε}_{0} L}$
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$\frac{q Q}{2 {πε}_{0} L}$ $\frac{q Q}{2 {πε}_{0} L}$
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$\frac{q Q}{6 {πε}_{0} L}$ $\frac{q Q}{6 {πε}_{0} L}$
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$- \frac{q Q}{6 {πε}_{0} L}$ $- \frac{q Q}{6 {πε}_{0} L}$

An electric dipole of moment $\vec{p}$ $\vec{p}$ is lying along a uniform electric field $\vec{E}$ $\vec{E}$ . The work done in rotating the dipole by 90 ° is :

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$\sqrt{2} p E$ $\sqrt{2} p E$
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$\frac{2 E}{2}$ $\frac{2 E}{2}$
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2pE
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pE

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery, the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

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decreases.
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does not change.
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becomes zero.
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increases.

The electric field vector in a region given by $\vec{E} = (3 \hat{i} + 4 y \hat{j}) V m^{- 1}$ $\vec{E} = (3 \hat{i} + 4 y \hat{j}) V m^{- 1}$ . Calculate the potential at (1m, 1m) taking potential at origin to be zero. [This question includes concepts from 12th syllabus]

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5V
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3V
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-1V
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-5V

Two-point charges each of charge +q are fixed at (+a, 0) and (-a, 0). Another positive point charge q placed at the origin is free to move along X-axis. The charge q at origin in equilibrium will have

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maximum force and minimum potential energy.
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minimum force and maximum potential energy.
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maximum force and maximum potential energy.
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minimum force and minimum potential energy.

Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density $σ$ $σ$ . They are brought in contact and separated. What will be the new surface charge densities on them?

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$σ_{1} = \frac{5}{6} σ_{}, σ_{2} = \frac{5}{6} σ$ $σ_{1} = \frac{5}{6} σ_{}, σ_{2} = \frac{5}{6} σ$
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$σ_{1} = \frac{5}{2} σ_{}, σ_{2} = \frac{5}{6} σ$ $σ_{1} = \frac{5}{2} σ_{}, σ_{2} = \frac{5}{6} σ$
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$σ_{1} = \frac{5}{2} σ_{}, σ_{2} = \frac{5}{3} σ$ $σ_{1} = \frac{5}{2} σ_{}, σ_{2} = \frac{5}{3} σ$
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$σ_{1} = \frac{5}{3} σ, σ_{2} = \frac{5}{6} σ$ $σ_{1} = \frac{5}{3} σ, σ_{2} = \frac{5}{6} σ$

Two identical capacitors C₁ and C₂ of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor C₁ using a battery of emf V volt. Now disconnecting a and b terminals, terminals b and c are connected. Due to this what will be the percentage loss of energy?

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75%
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0%
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50%
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25%

The electric potential V is given as a function of x (in metre) as $V = (x^{2} - 6 x + 5)$ $V = (x^{2} - 6 x + 5)$ . The electric field is zero at :

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x = 1m
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x = 2m
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x = 3m
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x = 6m

If n identical drops, each of capacitance C, coalesce to form a single big drop, the capacitance of the big drop will be

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$n^{3} C$ $n^{3} C$
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nC
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$n^{1 / 2} C$ $n^{1 / 2} C$
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$n^{1 / 3} C$ $n^{1 / 3} C$

Two charged conducting spheres of radii a and b are connected to each other by a wire. The ratio of electric fields at the surfaces of the two spheres is:

$(1) \frac{a}{b} (2) 1 (3) \frac{2 a}{b} (4) \frac{b}{a}$ $(1) \frac{a}{b} (2) 1 (3) \frac{2 a}{b} (4) \frac{b}{a}$

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1
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2
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3
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4

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm?

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1100 km²
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1130 km²
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1110 km²
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1105 km²

An electric field is given by $\vec{E} = (\hat{i} + 2 \hat{j} + \hat{k}) N / C$ $\vec{E} = (\hat{i} + 2 \hat{j} + \hat{k}) N / C$ . The work done in moving a 1 coulomb charge from $\vec{r_{A}} = (2 \hat{i} + 2 \hat{j)} m t o \vec{r_{B}} = (4 \hat{i} + \hat{j}) m$ $\vec{r_{A}} = (2 \hat{i} + 2 \hat{j)} m t o \vec{r_{B}} = (4 \hat{i} + \hat{j}) m$ is:

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8 J
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4 J
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-4 J
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Zero

Which of the following is incorrect about the electrostatic field lines?

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These can be never be closed curves
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On a conducting surface, the lines are perpendicular
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They can pass through a conductor
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If the lines are equispaced and parallel to one another, then the field is uniform

The conducting shells A and B are arranged as shown below. If the charge on the shell B is q then electric flux linked with the spherical Gaussian surface S is

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$\frac{q}{ε_{0}}$ $\frac{q}{ε_{0}}$
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$- \frac{q}{2 ε_{0}}$ $- \frac{q}{2 ε_{0}}$
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$- \frac{q}{ε_{0}}$ $- \frac{q}{ε_{0}}$
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$\frac{q}{2 ε_{0}}$ $\frac{q}{2 ε_{0}}$

A particle of mass 2 g and charge 1 $μ C$ $μ C$ is held at a distance of 1 m from a fixed charge of 1 mC. If the particle is released then its speed, when it is at a distance of 10 m from the fixed charge, is

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55 m/s
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100 m/s
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45 m/s
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90 m/s

The insulation property of air breaks down at $E = 3 \times 10^{6}$ $E = 3 \times 10^{6}$ V/m. The maximum charge that can be given to a sphere of diameter 5 m is approximately

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$2 \times 10^{- 5} C$ $2 \times 10^{- 5} C$
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$2 \times 10^{- 4} C$ $2 \times 10^{- 4} C$
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$2 \times 10^{- 3} C$ $2 \times 10^{- 3} C$
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$3 \times 10^{- 3} C$ $3 \times 10^{- 3} C$

A: A sensitive electric equipment can be saved from the electric field by enclosing it inside a metallic shell

R: Electric field inside a metallic shell is zero provided that the shell does not enclose any charge

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1 If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark
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2 If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark
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3 If Assertion is a true statement but Reason is false, them mark
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4 If both Assertion and Reason are false statements, then mark

A: The electric field lines are normal to a conducting surface.

R: A conducting surface is an equipotential surface in equilibrium.

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1 If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark
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2 If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark
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3 If Assertion is a true statement but Reason is false, them mark
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4 If both Assertion and Reason are false statements, then mark

A: If the separation between two small electric dipoles is doubled without changing their relative orientation, the force between them becomes one eight of the initial value.

R: On the equatorial position of an electric dipole the potential is non zero

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1 If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark
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2 If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark
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3 If Assertion is a true statement but Reason is false, them mark
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4 If both Assertion and Reason are false statements, then mark

Some equipotential surfaces are shown in figure. The electric field at points A, B and C are respectively :-

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1 V/cm, $\frac{1}{2}$ $\frac{1}{2}$ V/cm, 2 V/cm (all along +ve X-axis)
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1 V/cm, $\frac{1}{2}$ $\frac{1}{2}$ V/cm, 2 V/cm (all along -ve X-axis)
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$\frac{1}{2}$ $\frac{1}{2}$ V/cm, 1 V/cm, 2 V/cm (all along +ve X-axis)
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$\frac{1}{2}$ $\frac{1}{2}$ V/cm, 1 V/cm, 2 V/cm (all along -ve X-axis)

In a certain region of space with volume 0.2 m³, the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is:

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0.5 N/C
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1 N/C
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5 N/C
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zero

A short electric dipole has a dipole moment of $16 \times 10^{- 9} C m$ $16 \times 10^{- 9} C m$ . The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole situated on a line making an angle of $60 °$ $60 °$ with the dipole axis is :

$(\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} N m^{2} / C^{2})$ $(\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} N m^{2} / C^{2})$

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200 V
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400 V
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zero
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50 V

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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