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As promised in the post dated 26

^{th}August 2009, I give below a few typical multiple choice practice questions for AP Physics C students:
(1) The displacement current in the dielectric between the plates of a 0.1 μF capacitor is 1 mA. The potential difference between the plates of this capacitor must be changing at the rate of

(a) 10

^{4}volt/second
(b) 10

^{3}volt/second
(c) 10

^{2}volt/second
(d) 10 volt/second

(e) 1 volt/second

This is a simple question contrary to the impression of some of you at the first glance! Whether it is conduction current or displacement current, we have the usual expression for electric current (

*i*) as*i =*d

*Q/*d

*t*

Since the charge

*Q = CV*, we have*i = C*d*V/*d*t*
This gives d

*V/*d*t*=*i/C*= 1 mA /0.1 μF = 10^{–3}ampere^{ }/10^{–7}farad = 10^{4}volt/second [Option (a)].
(2) A 10 μF electrolytic capacitor connected to a 6 V power supply is fully charged. The displacement current flowing through the dielectric is

(a) 100 mA

(b) 10 mA

(c) 1mA

(d) 0.1 mA

(e) zero

The correct option is (e) since there is no charging current in the case of the fully charged capacitor.

(3) The electric and magnetic fields in a radio wave propagating along the z-direction are given respectively by

*E*

_{x}=

*E*

_{0}sin (

*kz–*

*ωt*) and

*B*

_{y}=

*B*

_{0}sin (

*kz–*

*ωt*)

where

*E*_{0}and*B*_{0}are the amplitudes,*k*is the magnitude of the wave vector (propagation vector) and*ω*is the angular frequency. How is*E*_{0}related to*B*_{0}?
(a)

*E*_{0}=*B*_{0}*k**ω*
(b)

*E*_{0}=*B*_{0}*k/**ω*
(c)

*E*_{0 }=*B*_{0}√(*k**ω*)
(d)

*E*_{0}=*B*_{0}
(e)

*E*_{0 }=*B*_{0}*ω/k*
The electric and magnetic fields in the wave are varying simple harmonically and the speed of the wave is given by

*c*=*ω/k*. [Note that the speed of a wave propagating in the z-direction is given by the ratio of coefficient of*t*to the coefficient of*z*. [See the post dated 9^{th}April 2009 ‘AP Physics B- Wave Motion (including sound)- Points to be Noted’ which you can access by clicking on the label ‘waves’ below this post.
Since

*E*=*Bc*we have*E*_{0 }=*B*_{0}*c = B*_{0}*ω/k*
(4) The charge on the plates of a capacitor of plate area 0.1 m

^{2}is 0.05 coulomb and is decreasing at the rate of 0.5 coulomb per second at an instant*t*during its discharge. What is the displacement current at the instant*t*?
(a) 0.005 A

(b) 0.05 A

(c) 0.5 A

(d) 5 A

(e) Data insufficient for arriving at an answer

Data is more than sufficient. The plate area given in the question is not required and it serves the purpose of a distraction. The displacement current is the time rate of change of charge (d

*Q*/d*t*) on the plates and is equal to 0.5 ampere [Option (c)].
[If the capacitor is an ideal parallel plate capacitor and you are required to find the magnitude of the

*displacement current density*, the answer is 0.5 A/0.1 m^{2}= 5 Am^{–2}]
(5) At a particular point in space and time, the magnetic field in a plane electromagnetic wave traveling in free space along the positive

*x*-direction is given by**B**= 3.1×10^{–8}**k**T (where**i**,**j**,**k**are unit vectors along the*x*,*y*,*z*directions as usual). The electric field at the same point at the same instant is
(a) – 1.033

**j**Vm^{–1}
(b) 9.3

**i**Vm^{–1}
(c) – 9.3

**j**Vm^{–1}
(d) 1.033

**k**Vm^{–1}
(e) 9.3

**j**Vm^{–1}
The magnitude of the electric field is given by ind icated by the unit vector

*E = Bc =*3.1×10^{–8 }×3×10^{8}Vm^{–1}= 9.3 Vm^{–1}. Since the wave is proceeding along the positive*x*-direction and the magnetic field vector is along the positive*z*-direction (as**k**), the electric field must be along the positive*y*-direction. [Note that the cross product vector**E****×B**should point along the direction of propagation].
Therefore,

**E**= 9.3**j**Vm^{–1}.
(6) A microwave signal traveling through the earth’s atmosphere has an electric field of amplitude 0.02 Vm

^{–1}. The intensity (power flow per unit area) of this signal is (Permittivity of free space,*ε*_{0 }= 8.85×10^{–12}C^{2 }N^{–1 }m^{–2})
(a) 5.31×10

^{–2}Wm^{–2}
(b) 5.31×10

^{–3}Wm^{–2}
(c) 5.31×10

^{–6}Wm^{–2}
(d) 5.31×10

^{–7}Wm^{–2}
(e) 5.31×10

^{–8}Wm^{–2}
The intensity (

*I*) which is the power flow through unit area (with the plane of the area perpendicular to the direction of propagation) is given by*I =*

*ε*

_{0 }

*c*

_{ }

*E*

_{m}

^{2}

*/*2 where

*c*is the speed of the signal and

*E*

_{m}is the amplitude of the electric field.

Therefore,

*I =*(8.85×10^{–12}×3×10^{8}×0.0004)*/*2 = 5.31×10^{–7}Wm^{–2}
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