1. An oil drop weighs 1.9 X 10-15 N. It is suspended in an electric field intensity of 6.0 X 103 N/C. a) What is the charge on the drop?
b) How many excess electrons does it carry?

2. A positively-charged oil drop weighs 6.4 X 10-13 N. An electric field intensity of 4.0 X 106 N/C suspends the drop.
a) What is the charge on the drop?
b) How many electrons is the drop missing?

3. If three more electrons were removed from the drop described in problem 2, what field would be needed to balance the drop?

4. A drop is falling in a Millikan oil drop apparatus when the electric field is off.
a) What are the forces acting on it ?
b) If it is falling at a constant velocity, what can be said about the forces acting on it?

5. a) What is the magnitude of the force acting on an electron in an electric field of 800 N/C?

b) A proton (mass:1.67 X 10-27 kg) is suspended at rest in a uniform electric field. Take into account gravity and determine E.

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Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge