Chapter 19: Electric Charges, Forces, and Fields Selected Solutions

Selected Solutions

13. (a) Let the x-axis be along the line of the three charges with the positive direction pointing from q₂ to q₃. Then, because opposite charges attract we have that F₂₁ points in the negative x-direction and F₂₃ points in the positive x-direction. When combined with Coulomb's law for the magnitudes, this gives

.

The net force on charge q₂, F₂, is the vector sum of these two forces:

Therefore, the net electrostatic force exerted on q₂ is 200 N towards q₃.

(b) Since , if d were tripled, the answer to part (a) would decrease by a factor of 1/3², or 1/9.

31. (a) The forces balance. So, the force due to the electric field must be opposite to that due to gravity. And, since the charge is negative, the electric field must be directed downward.

(b) Since the downward force due to gravity was balanced by the upward force do to the electric field, and since the charge on the object has now increased, the acceleration will be upward.

37. (a) The electric field lines begin at q₁ and q₃ and end at q₂. Also, electric field lines start at positive charges or at infinity and end at negative charges or at infinity. So, since q₂ is negative, q₁ and q₃ must be .

(b) q₁ has 8 lines leaving it. q₂ has 16 lines entering it. Since 8 is half of 16, and since the number of lines entering or leaving a charge is proportional to the magnitude of the charge, the magnitude of q₁ is one-half of q₂, or

(c) By the reasoning of part (b), the magnitude of q₃ is

Selected Solutions by David Reid, Eastern Michigan University. ©2002 by Prentice Hall, Inc.

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