Question

2. Suppose you have two small pith balls that are 6.5 cm apart and have equal charges of -27 nC? What is the magnitude of the repulsive force, in newtons, between the two pith balls?

3. Suppose you have two point charges each of 72 NC. How many millimeters apart must the two charges be to have a force of 1.05 N between them?

4. A free electron is suspended in an electric field near the surface of the Earth.

a. Give an expression for the magnitude of this electric field assuming the electron is stationary, in terms of the mass m_e and charge e of the electron, and the gravitational acceleration g.

b. What is the magnitude of this electric field, in N/C, assuming the electron is stationary?

c. (multiple choice) What does the small value for this field imply regarding the relative strength of the gravitational and electrostatic interaction?

(options) 1) The coulomb force is extraordinarily stronger than gravity.
2) The coulomb force is the same as gravity.
3) The coulomb force is extraordinarily weaker than gravity.
4) The coulomb force has nothing to do with gravity.
5) None of these.

5. Suppose you have a 3.9 μC charge and a 230 N/C electric field that points due east. What is the magnitude of the force, in newtons, exerted on the charge by the electric field?

Answer 1

Solution)(2) r = 6.5 cm = 6.5×10^(-2) m

Q1 = Q2 = - 27 nC = - 27×10^(-9) C

Force , F = ?

F = ((K)(|Q1|)(|Q2|))/(r^2)

K = 9×10^(9) Nm^2/C^2

F = (9×10^(9)×27×10^(-9)×27×10^(-9))/(6.5×10^(-2))^2

F = 15.5×10^(-4) N

(3) Q1 = Q2 = 72 nC = 72×10^(-9) C

F = 1.05 N

r = ?

F = ((K)(|Q1|)(|Q2|))/(r^2)

r^2 = ((K)(|Q1|)(|Q2|))/(F)

r^2 = (9×10^(9)×72×10^(-9)×72×10^(-9))/(1.05)

r^2 = (4443.428×10^(-8))

r = (4443.428×10^(-8))^(1/2)

r = 6.66×10^(-3) m

r = 6.66 mm

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