JEE Questions for Physics Electrostatics I Quiz 11 - MCQExams.com

The electric field and the potential of an electric dipole vary with distance r as

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  • 2)
    Physics-Electrostatics I-71138.png

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A dipole of electric dipole moment p is placed in a uniform electric field of strength E. If θ is the angle between positive directions of p and E, then the potential energy of the electric dipole is largest when θ is

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    Physics-Electrostatics I-71142.png

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  • Zero
A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x–y plane with its centre at (–a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8 C is placed on the a-axis from x = a/4 to x = 5a/4. Two point charges –7 C and 3 C are placed at (a/4, – a/4,and(–3 a/4, 3 a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a / 2, y = ± a / 2, z = ±a / 2. The electric flux through this cubical surface is
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    Physics-Electrostatics I-71147.png

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Total electric flux coming out of a unit positive charge put in air is

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    Physics-Electrostatics I-71152.png

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For a given surface the Gauss\'s law is stated as ∮ E. ds = 0. From this we can conclude that
  • E is necessarily zero on the surface
  • E is perpendicular to the surface at every point
  • The total flux through the surface is zero
  • The flux is only going out of the surface

Physics-Electrostatics I-71156.png
  • Zero
  • l2E
  • 4 l2E
  • 6 l2E
A point charge +q is placed at the centre of a cube of side L. The electric flux emerging from the cube is

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  • Zero

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A charge q is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is
  • Zero
  • 2)
    Physics-Electrostatics I-71161.png

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It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss\'s theorem because
  • Gauss's law fails in this case
  • This problem does not have spherical symmetry
  • Coulomb's law is more fundamental than Gauss's law
  • Spherical Gaussian surface will alter the dipole moment
According to Gauss\' theorem, electric field of an infinitely long straight wire is proportional to
  • r
  • 1/r2
  • 1/r3
  • 1/r
The diameter of each plate of an air capacitor is 4 cm. To make the capacity of this plate capacitor equal to that of 20 cm diameter sphere, the distance between the plates will be
  • 4 × 10–3 m
  • 1 × 10–3 m
  • 1 cm
  • 1 × 10–3 cm
Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1 m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is
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    Physics-Electrostatics I-71168.png

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The S.I. unit of electric flux is
  • Weber
  • Newton per coulomb
  • Volt x metre
  • Joule per coulomb
q1, q2, q3 and q4 are point charges located at points as shown in the figure and S is a spherical Gaussian surface of radius R. Which of the following is true according to the Gauss\'s law
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  • None of the above
Gauss\'s law should be invalid if
  • There were magnetic monopoles
  • The inverse square law were not exactly true
  • The velocity of light were not a universal constant
  • None of the above
A sphere of radius R has a uniform distribution of electric charge in its volume. At a distance x from its centre, for x < R, the electric field is directly proportional to

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    Physics-Electrostatics I-71179.png

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If the electric flux entering and leaving an enclosed surface respectively is ɸ1 and ɸ2 the electric charge inside the surface will be
  • (ɸ1 + ɸ2)ɛ0
  • (ɸ2 – ɸ1)ɛ0
  • (ɸ1 + ɸ/ ɛ0
  • (ɸ2 – ɸ/ ɛ0
What about Gauss\'s theorem is not incorrect?
  • It can be derived by using Coulomb's law
  • It is valid for conservative field obeys inverse square root law
  • Gauss theorem is not applicable in gravitation
  • (a) and (b) both
Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface the electric field will be due to
Physics-Electrostatics I-71184.png
  • q2
  • Only the positive charges
  • All the charges
  • + q1 and – q1
The electric flux for Gaussian surface A that enclose the charged particles in free space is (given q1 = –14 nC, q2 = 78.85 nC, q3 = –56 nC)
Physics-Electrostatics I-71185.png
  • 103Nm2C–1
  • 103 CN–1 m–2
  • 6.32 × 103Nm2 C–1
  • 6.32 × 103 CN–1 m–2
A point charge causes an electric flux of –1.0 × 103 Nm2 C–1 to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. If the radius of the Gaussian surface were three times, how much flux would pass through the surface
  • 3.0 × 103 Nm2/C
  • –1.0 × 103 Nm2/C
  • –3.0 × 103 Nm2/C
  • –2.0 × 103 Nm2/C
The electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm surrounding the total charge is 20 Vm. The flux over a concentric sphere of radius 20 cm will be
  • 20 Vm
  • 25 Vm
  • 40 Vm
  • 200 Vm
Which of the following is the correct statement of Gauss\'s law for electrostatics in a region of charge distribution in free space

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The electrostatic potential inside a charged spherical ball is given by ɸ = ar2 + b, where r is the distance from the centre; a, b are constants. Then the charge density inside the ball is
  • –24πaɛ0r
  • –6aɛ0r
  • –24πaɛ0
  • –6aɛ0
The potential energy of a charged parallel plate capacitor is U0, if a slab of dielectric constant k is inserted between the plates, then the new potential energy will be

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The energy of a charged capacitor is given by the expression (q = charge on the conductor and C = its capacity)

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    Physics-Electrostatics I-71201.png

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In a charged capacitor, the energy resides
  • The positive charges
  • Both the positive and negative charges
  • The field between the plates
  • Around the edge of the capacitor plates
If the circumference of a sphere is 2m, then capacitance of sphere in water would be
  • 2700 pF
  • 2760 pF
  • 2780 pF
  • 2800 pF
If the charge on a capacitor is increased by 2 coulomb, the energy stored in it increase by 21%. The original charge on the capacitor is
  • 10 C
  • 20 C
  • 30 C
  • 40 C
The potential gradient at which the dielectric of a condenser just gets punctured is called
  • Dielectric constant
  • Dielectric strength
  • Dielectric resistance
  • Dielectric number
A parallel plate condenser has a capacitance 50µF in air and 110µF when immersed in an oil. The dielectric constant \'k\' of the oil is
  • 0.45
  • 0.55
  • 1.10
  • 2.20
A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1 = 3 and thickness d/3 while the other one has dielectric constant k2 = 6 and thickness 2d/3. Capacitance of the capacitor is now
  • 45 pF
  • 40.5 pF
  • 20.25 pF
  • 1.8 pF
1000 small water drops each of radius r and charge q coalesce together to form one spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of
  • 1
  • 1000
  • 100
  • 10
Two large metal plates are placed parallel to each other. The inner surfaces of plates are charged by +σ and – σ (Coulomb/m2). The outer surfaces are neutral. The electric field is .... in the region between the plates and .... outside the plates.
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The capacitance of a spherical condenser is 1 µF. If the spacing between the two spheres is 1 mm, then the radius of the outer sphere is
  • 30 cm
  • 6 m
  • 5 cm
  • 3 m
A 500 µF capacitor is charged at a steady rate of 100 µC/ Second. The potential difference across the capacitor will be 10 V after an interval of
  • 5 sec
  • 25 sec
  • 20 sec
  • 50 sec
The capacity of a parallel plate capacitor increases with the
  • Decrease of its area
  • Increase of its distance
  • Increase of its area
  • None of the above
Two parallel plate of area A are separated by two different dielectrics as shown in figure. The net capacitance is
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    Physics-Electrostatics I-71221.png

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Can a metal be used as a medium for dielectric
  • Yes
  • No
  • Depends on its shape
  • Depends on dielectric
The respective radii of the two spheres of a spherical condenser are 12 cm and 9 cm. The dielectric constant of the medium between them is 6. The capacity of the condenser will be
  • 240 pF
  • 240 µF
  • 240 F
  • None of these
A capacitor of capacitance value 1 µF is charged to 30 V and the battery is then disconnected. If it is connected across a 2 µF capacitor, the energy lost by the system is
  • 300 µJ
  • 450 µJ
  • 225 µJ
  • 150 µJ
The energy stored in the capacitor as shown in the figure (a) is 4.5 × 10–6 J. If the battery is replaced by another capacitor of 900 pF as shown in figure (b), then the total energy of system is
Physics-Electrostatics I-71227.png
  • 4.5 × 10–6 J
  • 2.25 × 10–6 J
  • Zero
  • 9 × 10–6 J
A capacitor of capacity C has charge Q and stored energy is W. If the charge is increased to 2Q, the stored energy will be
  • 2W
  • W/2
  • 4W
  • W/4
Between the plates of a parallel plate condenser, a plate of thickness t1 and dielectric constant k1 is placed. In the rest of the space, there is another plate of thickness t2 and dielectric constant k2. The potential difference across the condenser will be

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A cylindrical capacitor has charge Q and length L. If both the charge and length of the capacitor are doubled, by keeping other parameters fixed, the energy stored in the capacitor
  • Remains same
  • Increases two times
  • Decreases two times
  • Increases four times
64 identical spheres of charge q and capacitance C each are combined to form a large sphere. The charge and capacitance of the large sphere is
  • 64q, C
  • 16q, 4C
  • 64 q, 4C
  • 16 q, 64 C
C, V, U and Q are capacitance, potential difference, energy stored and charge of parallel plate capacitor respectively. The quantities that increases when a dielectric slab is introduced between the plates without disconnecting the battery are
  • V and C
  • V and U
  • U and Q
  • V and Q
There is an air filled 1pF parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increase to 2pF. The dielectric constant of wax is
  • 2
  • 4
  • 6
  • 8
The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C0 and W0. If the air is replaced by glass (dielectric constant =between the plates, the capacity of the plates and the energy stored in it will respectively be

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Force of attraction between the plates of a parallel plate capacitor is

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