Electromagnetic theory miscellaneous

Electromagnetic theory miscellaneous

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Electromagnetic Theory

Electromagnetic theory miscellaneous
  1. Consider the following statements associated with the basic electrostatic properties of ideal conductors:
    1. The resultant field inside is zero.
    2. The net charge density in the interior is zero.
    3. Any net charge density in the interior is zero.
    4. The surface is always an equipotential.
    5. The field just outside is zero.
    Of the statements:








  1. View Hint View Answer Discuss in Forum

    Only statement 5 is not correct. The electic field, if any is radially outward at conducting surface governed by the relation, Dn = ρs, where Dn is the normal or radial component of electric displacement density and ρs is the surface charge density at the point of interest. Thus depending upon ρs, the field may be zero or non-zero.

    Correct Option: A

    Only statement 5 is not correct. The electic field, if any is radially outward at conducting surface governed by the relation, Dn = ρs, where Dn is the normal or radial component of electric displacement density and ρs is the surface charge density at the point of interest. Thus depending upon ρs, the field may be zero or non-zero.

  1. Three concentric spherical shells of radii R1, R2, R3 (R1 < R2 < R3) carry charges –1, –2 and 4 coulombs, respectively. The charge in coulombs on the inner and outer surfaces respectively, of the outermost shell is—








  1. View Hint View Answer Discuss in Forum

    As far as the conducting spherical shell of radius R3 is concerned, the charge contained in the volume enclosed by it is – 1 + (– 2) = – 3C. This will induce a charge of + 3C, on the inner surface and a charge of – 3C on the outer surface of this third shell. Also, net charge reside on the outer surface has a charge of + 3C and the outer surface has 4 – 3 = 1C.

    Correct Option: B

    As far as the conducting spherical shell of radius R3 is concerned, the charge contained in the volume enclosed by it is – 1 + (– 2) = – 3C. This will induce a charge of + 3C, on the inner surface and a charge of – 3C on the outer surface of this third shell. Also, net charge reside on the outer surface has a charge of + 3C and the outer surface has 4 – 3 = 1C.

  1. A magnetic field →B = ( →a x + 2 →a y· 4 →a z) exists at a point is →E, a test charge moving with a velocity, →v = v0 (3 →a x – →a y + 2 →a z) experiences no force at a certain point, the electric field at that point will be—








  1. View Hint View Answer Discuss in Forum

    B = →→→
    (ax+2ay - 4az)

    v = v0→→→
    (3ax-ay + 2az)

    Force on the charge q moving with velocity v in a magnetic field B is given by F = q.v
    =V0 →→→→→→
    (3ax-ay+az)×(ax+ay-4ax)

    =V0 →→→→→→
    (6ax-12ay+az+ 4ax+2ay-4ax)

    = v0→→
    (14ay + 7az)

    The electric force is FE = qE
    Thus, qv × B – qE=0
    or
    E = – v × B
    ∴ E = – v0→→
    (14ay + 7az)

    Correct Option: B

    B = →→→
    (ax+2ay - 4az)

    v = v0→→→
    (3ax-ay + 2az)

    Force on the charge q moving with velocity v in a magnetic field B is given by F = q.v
    =V0 →→→→→→
    (3ax-ay+az)×(ax+ay-4ax)

    =V0 →→→→→→
    (6ax-12ay+az+ 4ax+2ay-4ax)

    = v0→→
    (14ay + 7az)

    The electric force is FE = qE
    Thus, qv × B – qE=0
    or
    E = – v × B
    ∴ E = – v0→→
    (14ay + 7az)

  1. Two infinite parallel metal plates are charged with equal surface charge density of the same polarity. The electric field in the gap between the plates is—








  1. View Hint View Answer Discuss in Forum

    For an infinite sheet of charge having surface charge density ρs, the electric field on either side is ρs/ 2∈ and away from it (assuming charge to be positive). Therefore, in the region between the plates, the two fields are antiparallel and equal; and hence the electric field in the gap between the plates is zero.

    Correct Option: D

    For an infinite sheet of charge having surface charge density ρs, the electric field on either side is ρs/ 2∈ and away from it (assuming charge to be positive). Therefore, in the region between the plates, the two fields are antiparallel and equal; and hence the electric field in the gap between the plates is zero.

  1. Consider the following statements regarding field boundary conditions:
    1. The tangential component of electric field is continuous across the boundary between two dielectrics.
    2. The tangential component of electric field at a dielectric-conductor boundary is non-zero.
    3. The discontinuity in the normal component of the flux density at a dielectric-conductor boundary is equal to the surface charge density on the conductor.
    4. The normal component of the flux-density is continuous across the charge-free boundary between two dielectrics.
    Of these statements—








  1. View Hint View Answer Discuss in Forum

    The tangential components of electric field at the boundary are discontinuous by the surface current density; the surface currents flow only in high conductivity media; therefore, statement-1 is correct. There is no tangential component of electric field at any conductor surface, only the radial component exists, therefore, statement-2 is wrong. The displacement flux densities normal components at the boundary are discontinuous by surface charge density at the boundary. Therefore, statements 3 and 4 are correct.

    Correct Option: D

    The tangential components of electric field at the boundary are discontinuous by the surface current density; the surface currents flow only in high conductivity media; therefore, statement-1 is correct. There is no tangential component of electric field at any conductor surface, only the radial component exists, therefore, statement-2 is wrong. The displacement flux densities normal components at the boundary are discontinuous by surface charge density at the boundary. Therefore, statements 3 and 4 are correct.