Charging by Induction, Conduction, Frustration

HW: Q7,8, P19, 27, 45, 56
Questions:

7. The attraction of the negative charges to the positive charges is strong, but remember that the charges will also feel a repulsion force from being close to the other negative charges.

8. The leaves of the electroscope act like springs as they are forced apart. You can feel this force in a simple experiment - take a ruler and hold it against the edge of a table so that most of its length is hanging off the edge. As you push on the free end of the ruler, you will feel a force that is proportional to how far you push on the end (displacement).

Problems:
19. To figure out where the third charge must go, you have to observe that it can't be located between the charges, because the forces from each charge would be pointed in the SAME direction. You can then say that the third charge is located a distance x to the right of the negative charge.

F1 = kQ1Q/r1^2
F2 = kQ2Q/r2^2

substituting for r1, r2, and dividing by k:
Q1/(L + x)^2 = Q2/x^2

You can solve numerically, getting that x = .91 m or -.11 m. Since the -.11 m location puts the charge between the first two (which we previously said could not happen), only the x = .91 m solution works.

Important Information Alert!
Notice that in 27, 45, and 56, it is stated in the problem that electric field is constant. If you skim through the problem too quickly, you may miss this fact.

27. F = qE = ma so E = 7.10 x 10^-10 N/C. Since it is an electron, the electric field will be directed in the opposite direction of the acceleration, which would be to the south.

45. The weight of the water drop is balanced by the electric force.

mg - qE = 0.

To find m, you must use the volume of a sphere, 4/3*pi*r^3, times the density of water, 1000 kg/m^3.

You can then solve for Q, and then divide this value by the charge on an electron (1.6 x 10^-19 C) to find the number of electrons on the drop.

56. Draw a FBD: T + qE - mg = 0. Thus E = (mg - T)/Q = 1.06 x 10⁷ N/C. Since we assumed in our FBD that E was directed upwards, the negative answer means the field is actually directed downwards.