Exercise Solution
Problem 2
How many electrons make up the stated charge?
Use the charge per electron to find the number.
(30.0x10-6C)(1 e/1.602x10-19C) = 1.87x1014 electrons
Problem 3
Magnitude of the elctric force given q and d
Use Coulomb's law
F = kQ1Q2/r2 = k(1.602x10-19C)(26x1.602x10-19C)/d = 2.7 x 10-3 N
Problem 7
Force goes as 1/r squared, so place a variable by r in the equation and solve for the variable - this will be explained in class - simple algebra
variable = 1/(3)1/2
Radius = 8.45/(3)1/2 = 4.88 cm
Problem 13
This is an equilateral triangle so each angle is 60 degrees.
We let d represent the length of each side and Q represent each of the charges.
This problem is lengthy and a reasonable attempt will be given credit. You should recognize that the force at each vertex is a combination of two forces.
Let's look at vertex 1, at the top, numbered clockwise, so 2 is bottom left and 3 is bottom right.
F12 = kQ2/d2 ð F12x = F12cos(600) F12y = F12sin(600)
F13 = kQ2/d2 ð F13x = F13cos(600) F13y = F12sin(600)
F1x = F12x + F13x = 0
F1y = F12y + F13y
F1 = (F1x2 + F1y2)1/2
The direction is clearly up, 90 degrees from the horizontal.
The remaining vertices are addressed the same way
Question 17
Point A: net force on a positive test charge would be down, to the left, parallel to the nearby electric fields
Point B. up and to the right, parallel to the nearby electric field lines
Point C. 0
In order of decreasing field strength: A, B, C
Problem 16
Forces on point charges at the top right and bottom right corners of a square
Need to first establish a coordinate system - left bottom corner is a good choice to make distances positive.
Note that each of these charges will have a vertical component of force due to one charge and a diagonal component of force due to the other charge.
Same solution approach from here as for other vector problems
Problem 24
Proton released in electric field and it experiences a given electric force toward south.
Find magnitude and direction of the electric field.
We know the direction will be toward the south since that is the direction of the force
E = F/q
Problem 36
Two point charges, Q1 and Q2 of given magnitude, are separated by a given distance d. Q2 is to the right of Q1.
An electric field at a point is given. x is the distance from the point to Q1. How far apart are the charges?
In order for the field to be zero at the given point, we know that the magnitudes of the two fields must be equal (due to opposite sign charges)
E1 = E2 ð kQ1/x2 = kQ2/(x + d)2 and solve for x = 29 cm.
Problem 39
Both charges must be same sign so the electric fields oppose each other and can add to zero.
As above, the magnitudes must be equal.
E1 = E2 ð kQ1/(l3)2 = kQ2/(2l/3)2 solve for Q1/Q2 = 1/4