Two thin dielectric slabs of dielectric constants Kand K2 (K< K2) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field 'E' between the plates with distance 'd' as measured from plate P is correctly shown by: 

  
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  • 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 :

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

    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 with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ=90°, the torque and the potential energy of the dipole will respectively be

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

    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:

    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:

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

    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:

    The electric field vector in a region given by E=(3i^+4yj^)Vm-1. Calculate the potential at (1m, 1m) taking potential at origin to be zero. [This question includes concepts from 12th syllabus]

    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

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

    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?

    The electric potential V is given as a function of x (in metre) as V=(x2-6x+5). The electric field is zero at :

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

    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 ab2 13 2ab4 ba

    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?

    An electric field is given by E=(i^+2j^+k^) N/C. The work done in moving a 1 coulomb charge from rA=(2i^+2j)^m to rB=(4i^+j^)m is:

     

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

     

    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

    A particle of mass 2 g and charge 1 μC 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

    The insulation property of air breaks down at E = 3 × 106 V/m. The maximum charge that can be given to a sphere of diameter 5 m is approximately

    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

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

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

    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

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

    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:

    A short electric dipole has a dipole moment of 16×10-9 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° with the dipole axis is :

    (14πε0=9×109 N m2/C2)

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