1. Consider the two charges q1 and q2 arranged along the x axis as shown.
(a) There is a point on the x axis in the vicinity of these charges where the electric field is zero. This point is:
(x) to the left of q1
( ) between q1 and q2
( ) to the right of q2
The electric field points away from + charges and towards - charges. Between the two charges the field from each charge points right, so they can't cancel. Only to the left of q1 can the fields cancel.
(b) A third charge q3 = + 1 µC is to be placed on the x axis in a location so that the electric field at x = 1 is zero. At what x coordinate should q3 be placed?
It's a good idea to start by figuring out roughly where to place q3. It has to go to the left of x = 1, because the net field from q1 and q2 at x=1 is to the left so q3 has to generate a field going right. Because q2 is so much bigger than q1 and q3, q3 should be somewhere between q2 and x = 1. Let's say it is placed at a distance r from x = 1.
The first step is to write down the fields produced by the three charges. The net field is zero, so the field pointing left from q2 must be balanced by the two fields pointing right from q1 and q3. This gives the equation:
All the k's cancel, and so do all the factors of 10-6. This gives:
Re-arranging this gives:
This is the distance the charge is away from x = 1. We already figured out that it must be to the left of x = 1, so the x coordinate is:
x = 1 - r = 1 - 0.258 = 0.742 m
2. A hollow conducting sphere of radius 5 cm has a charge of +2.5 µC. Another hollow conducting sphere, of radius 10 cm, is placed around the first one, as shown in the diagram below. The larger sphere has a charge of -2.5 µC. The sphere have a common center.
(a) On the diagram, sketch the electric field.
(b) What is the electric field at a distance of 2.5 cm from the center?
Inside the spheres the electric field is zero.
(c) What is the electric field at a distance of 7.5 cm from the center?
This point is inside the outer sphere, so the outer sphere contributes nothing. The point is outside the inner sphere, however, so the field from the inner sphere is the same as the field from a point charge. The field can be calculated from the point charge equation:
E = kq / r2 = (8.99 x 109) (2.5 x 10-6) / (0.075)2 = 4.00 x 106 N / C directed away from the center.
(d) What is the electric field at a distance of 15 cm from the center?
This point is outside both spheres so the net field is calculated by treating both spheres as point charges. Because the charges are equal and opposite, the field produced by the inner sphere is exactly cancelled by the field from the outer sphere. The net field is zero.
3. (a) In the circuit above, through which resistor does the greatest current pass?
(x) 3 ohm
( ) 8 ohm
( ) 10 ohm
( ) 14 ohm (b) How much current does that resistor get?
The 3 ohm resistor gets all the current that comes out of the battery. This can be calculated using V = IR, with V the battery voltage and R the total resistance of the circuit. (Note that the 3 ohm resistor does not have V = 6 volts across it, so you can't use 6 = 3I to get the current.)
The set of four resistors has to be boiled down to one. The 10 ohm and 14 ohm resistors are in series, so they can be combined to 24 ohms. This is in parallel with the 8 ohm resistor; the equivalent resistance of 8 and 24 in parallel is 6 ohms. Add this to 3 to get 9 ohms, and that's the total resistance of the circuit.
I = V / R = 6 / 9 = 0.667 A
(c) In the circuit shown below, the two light bulbs are identical. With the switch open, which of these statements is true?
( ) A is brighter than B
( ) B is brighter than A
(x) the bulbs are equally bright
With the switch open everything is in series, so the current is the same everywhere.
(d) Now the switch is closed. Which of these statements is true?
(i) The brightness of bulb A (x) increases ( ) decreases ( ) stays the same
(ii) The brightness of bulb B ( ) increases (x) decreases ( ) stays the same
(iii) Thecurrent from the battery (x) increases ( ) decreases ( ) stays the same
The path through the switch represents a zero resistance path for the current, so with the switch closed the current all goes through the switch, rather than through B. B's brightness decreases (it's not even lit) but A's increases; the resistance in the circuit has been halved so the current is doubled.
(4) A single charged ion of helium (mass = 6.4 x 10-27 kg, charge = + 1.6 x 10-19 C) is accelerated from rest through a potential difference of 1000 V.
(a) What is the speed of the helium ion?
The kinetic energy of the ion after acceleration equals the potential energy it has at the start:
qV = 1/2 m v2
Solving for v gives v = 2.24 x 105 m/s
(b) The helium ion now enters a region containing a 1 T magnetic field. The magnetic field is into the page as shown in the sketch, and perpendicular to the path of the helium ion. Which way does the ion's path curve?
(x) towards the top of the page
( ) towards the bottom of the page
Your right hand should tell you that.
(c) Find the radius of its circular path.
r = mv / qB = 8.9 x 10-3 m.