## Pages

`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Friday, December 24, 2010

### AP Physics C - Multiple Choice Practice Questions involving Electric Flux and Gauss’s Law

Gauss’s law (Gauss theorem) in electrostatics is a very useful tool for calculating electrostatic field produced by a system of charges when the system has some symmetry. You might have realized how simple the calculation is if you choose a suitable Gaussian surface. You should definitely remember the statement of Gauss’s law:

“Electric flux through any closed surface S = Q0 where Q is the total charge enclosed by the surface S and ε0 is the permittivity of free space.”

Today we will discuss a few multiple choice practice questions involving electric flux and Gauss’s law:

(1) A closed cubical surface of side a contains an electric dipole of moment p. The total electric flux through the cubical surface is

(a) zero

(b) 8a2p

(c) 6a2p

(d) 8p/ε0a

(e) 6p/ε0a

The electric dipole contains positive and negative charges of equal magnitude. Therefore the total charge inside the cubical surface is zero. By Gauss’s law the total electric flux through the cubical surface is zero [Option (a)].

(2) A uniform electric field E given by E = 200 ĵ NC–1 exists in a certain region of space. The electric flux through a square of area A = 0.04 î m2 located in this field is (î and ĵ are unit vectors along the x-direction and y-direction respectively)

(a) 5000 Nm2C–1

(b) 5000 Nm2C–1

(c) 8 Nm2C–1

(d) 8 Nm2C–1

(e) zero

The area vector by convention is perpendicular to the plane of the area. Since the area vector is directed along the x-axis, you know that the square lies in the yz plane. Therefore, there is no electric flux passing through the square and the correct option is (e).

[In vector notation electric flux is the scalar product (dot product) of E and A. Therefore we have flux Ф = E.A = 200 ĵ . 0.04 î = 0].

(3) A cubical closed surface has N inward electric flux lines passing through one face and N outward electric flux lines passing through the opposite face as shown in the adjoining figure. The flux lines exhibit convergence and there are no flux lines passing through any other face. If the magnitude of the electric field at the centre of the cube is E, the total charge inside the cube is

(a) ε0N

(b) – ε0N

(c) 2ε0N

(d) – 2ε0N

(e) zero

Since the inward flux and the outward flux have the same value, the net flux through the closed surface is zero. By Gauss’s law, we have

0 = q0 where q is the total charge inside the closed surface.

Therefore q = 0.

(4) A point positive charge Q is placed at the centre of a cubical closed surface. The outward electric flux through one pair of opposite faces of the cube is

(a) zero

(b) Q0

(c) 2Q0

(d) Q/2ε0

(e) Q/3ε0

The total electric flux lines originating from the positive charge Q is Q0. Since these flux lines are distributed equally among the six faces (or three pairs of faces) of the cube, the flux through one pair of opposite faces of the cube is Q/3ε0.

(5) A hemispherical bowl of radius R is kept inverted on a horizontal table (so that the mouth of the bowl is in contact with the surface of the table). Assume that the surface of the table coincides with the XZ plane of a right handed Cartesian coordinate system. If a uniform electric field E directed along the negative z-direction exists in the entire region, what is the electric flux through the mouth of the hemispherical bowl?

(a) πR2E

(b) πR2E/3

(c) πR2E/3

(d) πR2E

(e) Zero

As the mouth of the bowl is lying in the XZ plane the electric field E (which is along the negative z-direction) is parallel to the plane of the mouth. The electric flux through the mouth of the bowl is therefore zero.

(6) A point charge Q is directly above the centre of a square of side a at distance a/2 from the centre (Fig.). The electric flux through the square is

(a) 2Q0

(b) Q0

(c) Q/2ε0

(d) Q/3ε0

(e) Q/6ε0

You may imagine the charge Q to be at the centre of a cube of side a. The total flux through all the six faces of the cube is Q0 by Gauss’s law. Now think of the square given in the question. Since it makes one face of the cube, the flux through it will be one sixth of the total flux. So the correct option is Q/6ε0.

Now, see some useful multiple choice questions (with solution) in this section here.

## Saturday, December 11, 2010

### AP Physics C - Multiple Choice Practice Questions on Transients in RC and RL circuits

“Nearly every man develops an idea, works it up to the point where it looks impossible, and then he gets discouraged. That’s not the place to become discouraged.”
– Thomas A. Edison
Electric transient phenomena associated with circuits containing capacitors and resistors (CR circuits) are important from theoretical as well as practical point of view and they find place in the AP Physics C course. Transient phenomena associated with circuits containing inductors and resistors (LR circuits) are not as important as those associated with CR circuits but you should not ignore them.
Today we will discuss some multiple choice practice questions in this section. It will be a good idea to get yourself equipped sufficiently for the exercise by referring to my on this site.
Here are the questions with their solution:

(1) In the circuit shown the the battery has emf E and internal resistance r. The switch S is closed at time t = 0. If the voltage across the capacitor is 90% of the emf E of the battery after time t’, what is the value of t’?
(a) RC ln10
(b) e–0.9/RC
(c) e0.9/RC
(d) (R+r)C ln(0.9)
(e) (R+r)C ln10
When a capacitor of capacitance C farad is charged by connecting it in series with a resistance R ohm and a battery of emf V volt, the charge Q coulomb on the capacitor after a time t seconds is given by
Q =Q0 (1 e–t/RC) where Q0 is the final (maximum) charge and ‘e’ is the base of natural logarithms.
Since the voltage V across the capacitor is Q/C, it follows that
V =V0 (1 e–t/RC) where V0 is the final (maximum) voltage which is the emf E of the battery.
Therefore we have
0.9 E = E[1 e–t’/(R+r)C] since the total resistance in the circuit is R+r.
This gives e–t/(R+r)C = 0.1
Or, et/(R+r)C = 10 from which t’ = (R+r)C ln10
(2) The adjoining figure shows a resistance R connected in series with an inductance L and a battery of emf E through a switch S. The internal resistance of the battery is negligible. Which one of following graphs represents the nature of variation of the potential drop V across the resistance with time?

If an inductor of inductance L and a resistor of resistance R are connected in series with a battery, the current I in the circuit increases exponentially with time and is given by
I = I0 (1 e–Rt/L) where I0 is the final (maximum) current in the circuit.
This shows that the current at time t = 0 (immediately after closing the switch) is zero. As time t increases, the current increases exponentially and reaches the final maximum value I0 (= E/R) when t becomes infinite.
As V = IR the potential drop V across R also increases exponentially with time as indicated by graph (a).
(3) A capacitor C charged to V0 volt is discharged through the series combination of two resistances R ohm and 2R ohm by closing the switch S (Fig.). The base of natural logarithms is e. What is the current in the circuit when 3RC seconds are elapsed after closing the switch S?
(a) V0/(RC)
(b) V0/(3RC)
(c) V0/(3R)
(d) V0/(RCe)
(e) V0/(3Re)
The time constant of the circuit is 3RC and hence the voltage across the capacitor after 3RC seconds must be V0/e. You can write this if you remember how the time constant of an ‘RC circuit’ is defined.
The discharge current at this instant is equal to (V0/e)/(3R) = V0/(3Re).
[The initial charge Q0 on the capacitor decays exponentially with time t as given by the equation, Q = Q0 e–t/3RC where Q is the charge at time t. The discharge current at time t is given by
I = dQ/dt = (Q0/3RC) e–t/3RC
The negative sign just shows that the discharge current is opposite to the charging current. Ignoring the negative sign, the discharge current when t = 3RC is Q0/3RCe.
But Q0/C = V0, the initial voltage across the capacitor.
Therefore I = V0/(3Re)].
(4) In question no.3 above, the voltage across the resistance R and the current in it immediately after closing the switch S are respectively
(a) 3V0, V0/(3R)
(b) 2V0/3, zero
(c) V0/3, V0/(3R)
(d) zero, zero
(e) Unpredictable
This is a very simple question. But some of you will have some doubts. Initially the voltage across the capacitor is V0 and the total resistance connected across this voltage is 3R. Therefore, voltage drop across the resistor of value R must be V0/3.
[Note that the initial discharge current I0 is given by
I0 = dQ/dt = (Q0/3RC) e0, appropriate to t = 0.
Ignoring the negative sign, I0 = (Q0/3RC) = V0/3R since Q0/C = V0
The initial voltage across R is I0R = V0/3].