a Explanation: In a dielectric-dielectric boundary, if a free surface charge density exists at the interface, then the normal components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2

a
Explanation: In a dielectric-dielectric boundary, if a free surface charge density exists at
the interface, then the normal components of the electric flux density are discontinuous
at the boundary, which means Dn1 = Dn2

b Explanation: The resistor will absorb power and dissipate it in the form of heat energy. The potential between two points across a resistor will be negative

b
Explanation: The resistor will absorb power and dissipate it in the form of heat energy.
The potential between two points across a resistor will be negative

a Explanation: For a distant point P, the R1 and R2 will approximately be equal. R1 = R2 = r, where r is the distance between P and the midpoint of the two charges. Thus they are in geometric progression, R1R2=r2 Now, r2 = 5 x 7 = 35. We get r = 5.91cm

a
Explanation: For a distant point P, the R1 and R2 will approximately be equal.
R1 = R2 = r, where r is the distance between P and the midpoint of the two charges.
Thus they are in geometric progression, R1R2=r2
Now, r2 = 5 x 7 = 35. We get r = 5.91cm

a Explanation: The potential due the dipole is given by, V = m cos θ/(4πεr 2 ). When the angle becomes perpendicular (θ = 90). The potential becomes zero since cos 90 will become zero.

a
Explanation: The potential due the dipole is given by, V = m cos θ/(4πεr
2
). When the
angle becomes perpendicular (θ = 90). The potential becomes zero since cos 90 will
become zero.

## The nnormal component of the electric flux density is always discontinuous at the interface. State True/False.

a Explanation: In a dielectric-dielectric boundary, if a free surface charge density exists at the interface, then the normal components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2

a

See lessExplanation: In a dielectric-dielectric boundary, if a free surface charge density exists at

the interface, then the normal components of the electric flux density are discontinuous

at the boundary, which means Dn1 = Dn2

## The potential taken between two points across a resistor will be

b Explanation: The resistor will absorb power and dissipate it in the form of heat energy. The potential between two points across a resistor will be negative

b

See lessExplanation: The resistor will absorb power and dissipate it in the form of heat energy.

The potential between two points across a resistor will be negative

## The bound charge density and free charge density are 12 and 6 units respectively. Calculate the susceptibility

c Explanation: The electric susceptibility is given by, χe = Bound free density/Free charge density. χe = 12/6 = 2. It has no unit

c

See lessExplanation: The electric susceptibility is given by, χe = Bound free density/Free charge

density. χe = 12/6 = 2. It has no unit

## Compute the power consumed by a material with current density 15 units in an area of 100 units. The potential measured across the material is 20V.

c Explanation: Power is given by, P= V X I, where I = J X A is the current. Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.

c

See lessExplanation: Power is given by, P= V X I, where I = J X A is the current.

Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.

## Calculate the energy in an electric field with flux density 6 units and field intensity of 4 units.

a Explanation: The energy in an electric field is given by, W = 0.5 x D x E, where D = 6 and E = 4. We get W = 0.5 x 6 x 4 = 12 units.

a

See lessExplanation: The energy in an electric field is given by, W = 0.5 x D x E, where D = 6

and E = 4. We get W = 0.5 x 6 x 4 = 12 units.

## Calculate the distance between two charges of 4C forming a dipole, with a dipole moment of 6 units

b Explanation: The dipole moment is given by, M = Q x d. To get d, we rearrange the formula d = M/Q = 6/4 = 1.5units

b

See lessExplanation: The dipole moment is given by, M = Q x d. To get d, we rearrange the

formula d = M/Q = 6/4 = 1.5units

## For two charges 3C and -3C separated by 1cm and are located at distances 5cm and 7cm respectively from the point P, then the distance between their midpoint and the point P will be

a Explanation: For a distant point P, the R1 and R2 will approximately be equal. R1 = R2 = r, where r is the distance between P and the midpoint of the two charges. Thus they are in geometric progression, R1R2=r2 Now, r2 = 5 x 7 = 35. We get r = 5.91cm

a

See lessExplanation: For a distant point P, the R1 and R2 will approximately be equal.

R1 = R2 = r, where r is the distance between P and the midpoint of the two charges.

Thus they are in geometric progression, R1R2=r2

Now, r2 = 5 x 7 = 35. We get r = 5.91cm

## Find the potential due the dipole when the angle subtended by the two charges at the point P is perpendicular.

a Explanation: The potential due the dipole is given by, V = m cos θ/(4πεr 2 ). When the angle becomes perpendicular (θ = 90). The potential becomes zero since cos 90 will become zero.

a

See lessExplanation: The potential due the dipole is given by, V = m cos θ/(4πεr

2

). When the

angle becomes perpendicular (θ = 90). The potential becomes zero since cos 90 will

become zero.

## Calculate the dipole moment of a dipole with equal charges 2C and -2C separated by a distance of 2cm.

b Explanation: The dipole moment of charge 2C and distance 2cm will be, M = Q x d. Thus, M = 2 x 0.02 = 0.04 C-m.

b

See lessExplanation: The dipole moment of charge 2C and distance 2cm will be,

M = Q x d. Thus, M = 2 x 0.02 = 0.04 C-m.

## Compute the power consumed by a material with current density 15 units in an area of 100 units. The potential measured across the material is 20V

c Explanation: Power is given by, P= V X I, where I = J X A is the current. Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.

Explanation: Power is given by, P= V X I, where I = J X A is the current.

Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.