Practice Problems: The Basics of Electrostatics Solutions
1. (easy) A point charge (q1) has a magnitude of 3x10-6C. A second charge (q2) has a magnitude of -1.5x10-6C and is located 0.12m from the first charge. Determine the electrostatic force each charge exerts on the other.
F = kqq/r2
F = (9x109)(3x10-6)(1.5x10-6)/(0.12)2
F = 2.81 N (Attractive)
2. (easy) Describe the changes to the magnitude and direction of the force on one of the charges in an electric dipole when the distance between the charges increase.
The magnitude of the force on each charge will decrease on the inverse square with distance, but the direction of the force will stay the same.
3. (easy) The picture you see here is of a device called an "electroscope". It is used to detect the presence of excess charge. The top of the electroscope is made of a metal ball (a conductor). The ball is attached to two very lightweight metal "leaves" by means of a conducting shaft. The leaves are protected from the outside by a glass case. If a positively charged rod was held close to the top of the metal ball of the electroscope, the leaves would spread apart due to an excess of what type of charge in the leaves?
Some of the electrons in the leaves would rush to the top of the apparatus as they are attracted to the rod. Thus, there is an excess of positive charge left behind in the leaves.
4. (easy) If the electroscope in the previous question was charged by conduction to have excess electrons on the leaves, the leaves would spread out as shown in the picture. If a positively charged object was then brought in close to the metal ball on the top of the electroscope would the leaves spread further or would they come closer together?
The leaves would get closer because some of the excess electrons in the leaves would rush up the stem. The charge on the leaves would decrease and they would collapse (at least somewhat).
5. (easy) Now assume that the electroscope from question #3 has no excess charge. If a positively charged rod is then held close to the left side of the metal ball while the right side of the ball is grounded, the leaves will remain separated after the ground and the rod are removed (in that order). If a negatively charged rod was then placed next to the metal ball would the leaves spread apart more or come closer together?
They would spread further apart. The electroscope was charged negatively by induction at first. When the negative rod is then introduced some electrons in the metal ball would run down the stem making the leaves even more negative and exert greater separation force.
6. (easy) Evaluate the sketch below, assuming that the dot in the center of each line is the same charge for each arrangement. Determine which central charge has the greatest net force acting on it. Additionally, comment on the net force on the central charge of the other two arrangements. The arrangements do not interact with each other and the charges are evenly spaced. Finally, determine the direction of the net force on the central charge if it is positive.
According to Coulomb's Law, the top arrangement is the only one that will produce a net force on the central charge. The middle and lower arrangements produce a net force of zero on the central charge. If the central charge is positive the top arrangement will exert a force on it that acts to the right.
7. (moderate) Four charges are located on the corners of a square. The sides of the square have a length of 0.05 m. The upper left corner has a charge of q. The upper right corner has a charge of -q. The lower left corner has a charge of 2q. The lower right corner has a charge of -2q. If the magnitude of q is 1.0x10-7C, find the magnitude of the force exerted on the charge at the lower left corner of this system.
Number the charges: lower left= 1, upper left= 2, upper right=3, lower right=4
Distances: use a = 0.05 m
r1,2= a, r1,3 = (√2)a, r1,4 = a
F1,2 = k(2q)(q)/a2 = 0.072 N(repulsive)
F1,3 = k(2q)(q)/(√2a)2 = 0.036 N (attractive, at a 45° angle)
F1,4 = k(2q)(2q)/a2 = 0.144 N (attractive)
ΣFx = F1,3cos(45) + F1,4 = 0.036cos45 + 0.144 = 0.17 N
ΣFy = F1,3sin(45) - F1,2 = -0.05 N
F = 0.18 N
8. (moderate) Two identical charges (mass = 1.0 x10-6 kg) are at rest on the surface of a hemispherical bowl of radius R = 0.25 m. Three forces act on each particle: The normal force, the weight force, and the repulsive force they exert on each other. The normal force acting on either charge (caused by the surface of the bowl) is at a 60° angle to the horizontal. Find the charge on each particle.
The normal force acts radially toward the center of the hemisphere.
The electric force acts to the left on the charge as shown, but to the right on the other charge.
The weight force acts downward.
Use +y up and +x to the right.
ΣFy = 0
Nsin60 - mg = 0
N = mg/sin60
ΣFx = 0
Ncos60 - FE = 0
Ncos60 = kqq/R2
(mg/sin60)cos60 = kqq/R2
1.0x10-6(9.8)/tan60 = 9x109(q2)/(0.25)2
q = 6.3x10-9 C
9. (hard) Two point charges are located on the x axis. They are both positive, but the one located at x = 0 has a charge of q while the one located at x = L has a charge of 4q. If a third charge is placed on the x axis in between the two charges so that the net force on ANY of the charges is zero, determine the magnitude of the third charge and its location.
Call the third charge q3
F3 = k[(q)(q3)/x2 - (4q)(q3)/(L - x)2] = 0
The terms inside the bracket must be equal to zero.
Solve for x:
(q)(q3)/x2 - (4q)(q3)/(L - x)2 = 0
1/x2 = 4/(L - x)2
3x2 + 2xL - L2 = 0
Use the quadratic formula to show that x = L/3
To find the charge of q3 set up a force equation for the charge located at x=0.
Fq = k[(q)(q3)/(L/3)2 - (q)(4q)/(L)2] = 0
The L cancels yielding a magnitude of q3 = -(4/9)q
Using these values, the larger charge, 4q, is also in equilibrium.