k q1 q2 / 12 = k q1 q3 / x2
So q2 = q3 / x2 and x2 = q3 / q2 = 4 / 1 = 4
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:
( ) to the left of q1
(x) between q1 and q2
( ) to the right of q2
The electric field points away from + charges and towards - charges. Both these charges are positive. To the left of q1, the electric fields from the two charges both point left, so they can't cancel. To the right of q2, the electric fields both point right, so again they can't cancel. Between the charges, one field points right and one points left. There will be some point between them, then, where the fields are equal and opposite, giving a net field of zero.
(b) A third charge q3 = + 4 C is to be placed on the x axis in a location so that the total force on q1 is zero. At what x coordinate should q3 be located?
It's a good idea to start by figuring out which side of q1 to place q3. It has to go on the left, so q1 will have one force going left and another going right. So, let's say that q3 is a distance x to the left of q1. The force it exerts on q1 must be equal and opposite to the force q2 exerts on q1. Setting these forces equal gives:
k q1 q2 / 12 = k q1 q3 / x2
So q2 = q3 / x2 and x2 = q3 / q2 = 4 / 1 = 4
This gives x = 2. The third charge must be placed 2 m to the left of q1, at x = -2.0 m.
2. A charge Q = +4.0 C is spread uniformly over a spherical shell of radius 0.1 m.
(a) What is the electric field at these two positions (give magnitude and direction):
i) at r = 0.8 m from the center of the sphere
Outside the sphere, E is the same as that from a point charge. E = kQ / r2 = (8.99 x 109) (4 x 10-6) / 0.82 = 5.62 x 104 N/C, directed away from the sphere.
ii) at r = 0.05 m from the center of the sphere
Inside a sphere like this, E = 0.
(b) How much work must be done to bring a charge q1 = 1.0 C from very far away to the surface of the sphere?
W = PE = U
Uo = kqQ / r where r is very large, so Uo = 0.
Uf = kqQ / r = (8.99 x 109) (1 x 10-6) (4 x 10-6) / 0.1 = 0.360 J
360 J of work must be done to bring the charge to the surface of the sphere.
(c) How much additional work must be done to move q1 from the surface of the sphere to the center?
Inside the sphere there are no forces to work against, so no additional work is necessary.
3. In the circuit below, the direction of the current in each branch is shown, but is not necessarily correct.
(a) The three currents have values of:
To solve for the currents, apply Kirchoff's rules. With the directions as shown on the diagram, applying the junction rule gives: I1 + I2 + I3 = 0
Applying the loop rule to the loop on the left gives: +4 - 2I1 - 12 = 0 so 2I1 = -8 so I1 = -4 A
Applying the loop rule to the loop on the right gives: +12 +3I2 -6 = 0 so 3I2 = -6 so I2 = -2 A
Plugging these back into the first equation gives: I3 = -I1 -I2 = +6 A
(b) The directions can be figured out by looking at the signs.
The sign on I1 is negative, so it goes opposite to the direction shown.
The sign on I2 is negative, so it goes opposite to the direction shown.
The sign on I3 is positive, so it is as shown on the diagram.
(c) The power dissipated in the 2 resistor is 32 W.
The easiest way to get that is to use: P = I12 R = 42 x 2 = 32 W
(4) The diagram shows an electron entering the region between a pair of parallel plates. There is a potential difference of 100 V across the plates, as shown; the plates are separated by d = 0.01 m.
(a) On the diagram, draw the electric field and sketch a reasonable path for the electron to follow in the region between the plates.
The electric field goes from + to -, so it is directed down. The electron is deflected up, following a parabolic path.
(b) To ensure that the electron passes between the plates without being deflected, a magnetic field is introduced in the region between the plates. If the electron speed is 1.2 x 10^5 m/s, what is the magnitude of the required magnetic field?
To pass through undeflected, the force exerted on the electron by the magnetic field must be equal and opposite to the force exerted by the electric field. In other words, qE = qvB, so B = E / v. All we need to do is find the magnitude of the electric field. Between parallel plates, the electric field is uniform and is given by:
E = V / d = 100 / 0.01 = 10000 V / m
This gives a magnetic field of B = E / v = 10000 / 1.2 x 105 = 8.33 x 10-2 T
(c) The direction of the magnetic field introduced in (b) should be:
into the paper
The force exerted on the electron by the electric field is up, meaning that the force exerted on the electron by the magnetic field is down. Also, remember that your right hand lies to you for a negative charge. If you do the right-hand rule with a velocity to the right and a force down, the right hand says the field is out of the page. Flipping this because it's a negative charge, the field must be into the page.