Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
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An electric dipole with dipole moment 4 × 10^−9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10^4 N C^−1.
Electric dipole moment, p = 4 × 10−9 C m Angle made by p with a uniform electric field, θ = 30° Electric field, E = 5 × 104 N C−1 Torque acting...
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How is the strength of the magnetic field at a point near a wire related to the strength of the electric current flowing in the wire?
The magnitude of magnetic field is directly proportional to the strength of the electric current flowing in the wire.
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A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region.
(c) : Magnetic field applied parallel to motion of electron exerts no force on it as θ = 0 and force = Bevsinθ = zero Electric field opposes motion of electron which...
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An electric dipole is placed in a uniform electric field.
(a) is always zero
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A point charge q is rotated along a circle in the electric field generated by another point charge Q.
The correct answer is (a) zero
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An electric dipole is placed in an electric field generated by a point charge.
(d) The torque on the dipole due to the field may be zero.
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Mark the correct options: (a) Gauss's law is valid only for symmetrical charge distributions.
(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.
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The energy density in the electric field created by a point charge falls off with the distance from the point charge as
The correct answer is (d) 1/r4.
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A particle moves in a region having a uniform magnetic field and a parallel, uniform electric field.
(d) a helix with nonuniform pitch.
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A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r
Charge enclosed by the cylindrical surface = λl By Gauss Theorem, electric flux = 1/εo × charge enclosed. = 1/εo(λl)
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