Charging Into The Field - Electric field and 2D Electrostatics

Problems 12, 13, 18, 22, 26, 29

12. Remember the trick using your thumbs to find the direction of the force - same distance for all three, and a 60 degree angle for each force. Don't forget to find the direction - 83.8 N directed AWAY from the center of the triangle.

13. Again, this is not actually as tedious as it could be since it is a square and the charges are all the same, so your answer will end up being the same for each charge. You must use the Pythagorean theorem to find the distance of each charge from the charge on the opposite corner of the square (1.41 m).

For each charge, you will end up with something like Fnetx = F1 + F2 sin 45, and Fnety = F3 + F2 cos 45. (The subscripts will be different for each charge.) The final answer should be 6.2 x 10⁵ N directed away from the center of the square.

18. A little mix of Newton's 2nd and electrostatics here - the FBD of the electron should have kQ1Q2/r^2 directed towards the center of the circular orbit. The acceleration is v^2/R.

Q1 = Q2 = e (the fundamental charge), so ke^2/R^2 = mv^2/R.

Thus R = ke^2/mv^2 = 2.2 x 10^-10 m. It is interesting that this is not actually so far off from the actual size of an atom - you have now used basic physics to derive this fact! Not bad, eh?

22. This is a basic application of the definition of electric field: E = F/q. The magnitude comes from rearranging this definition: F = qE = 5.6 x 10^-16 N. Since it is an electron, which has a negative charge, the force will be directed towards the West, opposite the field.

26. This problem is very much like the one we did in class during 7th period - pretend that there is a small test charge at the midpoint of the two charges. Both E1 and E2 are directed to the left (or away from the positive charge) so their magnitudes should be added together. The net field (using the electric field for a point charge E = kQ/R^2) is 3.2 x 10⁸ N/C.

29. We need to use the SAME concept from the other problems in 2D to solve this problem. Unfortunately it's all algebraic, but fear not. The key is to start with a good diagram. I'm going to take mine from the nice folks at the textbook company:


In the diagram, E_- represents the electric field direction from the negative charge, and the other represents the direction for the positive field.

The distance from the origin is x, so the distance from x to each charge is (x^2 + a^2)^(1/2).

The other trick is to find the angle, and THEN find the sine of the angle, but you don't actually have to break that up into two steps. We know from trigonometric ratios that Sin (theta) = opposite side/hypotenuse. The hypotenuse is (x^2 + a^2)^(1/2) and the opposite side of the angle in the right triangle drawn by E_- is a. this means that sin(theta) = a/(x^2 + a^2)^(1/2).

The net electric field in x will be zero, since the charges are the same. In the y-direction, you should get that E = 2*E_- sin(theta).

Substituting our expression for the sine and simplifying, E = 2kQa/(a^2 + x^2)^3/2