Review for exam #1
Chapter 21 #8
In Fig. 21-23, four particles form a square. The charges are and . (a) What is if the net electrostatic force on particles 1 and 4 is zero? (b) Is there any value of q that makes the net electrostatic force on each of the four particles zero? Explain.
8. For ease of presentation (of the computations below) we assume Q > 0 and q < 0 (although the final result does not depend on this particular choice).
(a) The x-component of the force experienced by q1 = Q is
which (upon requiring F1x = 0) leads to , or
(b) The y-component of the net force on q2 = q is
which (if we demand F2y = 0) leads to . The result is inconsistent with that obtained in part (a). Thus, we are unable to construct an equilibrium configuration with this geometry, where the only forces present are given by Eq. 21-1.
Chapter 21 #28
Two tiny, spherical water drops, with identical charges of , have a center-to-center separation of 1.00 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?
28. (a) Eq. 21-1 gives
(b) If n is the number of excess electrons (of charge –e each) on each drop then
Chapter 22 #7
Two particles are fixed to an x axis: particle 1 of charge C at cm and particle 2 of charge at cm. At what coordinate on the axis is the net electric field produced by the particles equal to zero?
7. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge q2= 4.00 q1located at x2 = 70 cm has a greater magnitude than q1= 2.1 108 C located at x1 = 20 cm, a point of zero field must be closer to q1 than to q2. It must be to the left of q1.
Let x be the coordinate of P, the point where the field vanishes. Then, the total electric field at P is given by
.
If the field is to vanish, then
Taking the square root of both sides, noting that |q2|/|q1| = 4, we obtain
.
Choosing –2.0 for consistency, the value of x is found to be x = 30 cm.
Chapter 22 #9
In Fig. 22-32 , the four particles form a square of edge length cm and have charges . In unit-vector notation, what net electric field do the particles produce at the square's center?
9. The x component of the electric field at the center of the square is given by
Similarly, the y component of the electric field is
Thus, the electric field at the center of the square is
Chapter 22 #11
Figure 22-34 shows two charged particles on an x axis: at and at . What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the net electric field produced at point P at ?
11. (a) The vertical components of the individual fields (due to the two charges) cancel, by symmetry. Using d = 3.00 m and y = 4.00 m, the horizontal components (both pointing to the –x direction) add to give a magnitude of
.
(b) The net electric field points in the –x direction, or 180 counterclockwise from the +x axis.
Chapter 23 #2
An electric field given by pierces a Gaussian cube of edge length and positioned as shown in Fig. 23-5. (The magnitude is in Newton per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube?
2. We use and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m.
(a) On the top face of the cube y = 2.0 m and . Therefore, we have
. Thus the flux is
(b) On the bottom face of the cube y = 0 and . Therefore, we have
. Thus, the flux is
(c) On the left face of the cube . So
(d) On the back face of the cube . But since has no z component . Thus, = 0.
(e) We now have to add the flux through all six faces. One can easily verify that the flux through the front face is zero, while that through the right face is the opposite of that through the left one, or +16 N·m2/C. Thus the net flux through the cube is
= (–72 + 24 – 16 + 0 + 0 + 16) N·m2/C = –48 N·m2/C.