Final answer:
The charge on the suspended oil drop is approximately 8.086 x 10^-18 C, determined by balancing the gravitational force and the electric force. Dividing this charge by the fundamental charge of an electron gives approximately 51 extra electrons.
Step-by-step explanation:
To determine the charge on the oil drop, we use the electrical force and gravitational force equilibrium condition. The force due to the electric field (E) is F = qE, where q is the charge to be calculated and E is the electric field strength. The electric field (E) is calculated by dividing the applied potential difference (V) by the distance (d) between the plates, which gives us E = V/d. The gravitational force on the drop, which is its weight (W), is given by W = mg, where m is the mass of the drop and g is the acceleration due to gravity.
Since the oil drop is suspended and stationary, the forces must balance, so qE = mg. Substituting in the electric field, we have q = mgd/V. Using the given values, mass m = 4.89 x 10^-15 kg, gravitational acceleration g = 9.8 m/s^2, distance d = 0.05 m, and potential difference V = 3000 V, this gives:
q = (4.89 x 10^-15 kg)(9.8 m/s^2)(0.05 m) / (3000 V)
q ≈ 8.086 x 10^-18 C
To find the number of extra electrons, we divide the charge of the oil drop by the fundamental charge of an electron, which is approximately 1.6 x 10^-19 C. This gives us:
Number of electrons ≈ q/e ≈ (8.086 x 10^-18 C) / (1.6 x 10^-19 C/electron)
The number of electrons is approximately 50.54, which we round to the nearest whole number, since we can't have a fraction of an electron.