Answer:
Step-by-step explanation:
The potential difference required to maintain the equilibrium of a charged oil drop in a uniform electric field can be calculated using the following equation:
V = (mg)/(qE)
where V is the potential difference, m is the mass of the oil drop, g is the acceleration due to gravity, q is the charge on the oil drop, and E is the electric field strength.
We are given the mass of the oil drop (m = 1.31x10^-14 kg), the charge on the oil drop (q = 6.4x10^-19 C), and the separation between the plates (d = 10 mm = 0.01 m).
The electric field strength between two parallel plates separated by a distance d and with a potential difference V applied between them is given by:
E = V/d
Substituting this expression for E into the equation for V, we get:
V = (mg)/(qE) = (mgd)/(qV)
Rearranging this equation, we obtain:
V^2 = (mgd)/q
Substituting the given values, we get:
V^2 = [(1.31x10^-14 kg)(9.81 m/s^2)(0.01 m)]/(6.4x10^-19 C)
V^2 = 2.02x10^-5 V^2
Taking the square root of both sides, we obtain:
V = 0.0045 V
Therefore, the potential difference required to maintain the equilibrium of the charged oil drop is approximately 0.0045 volts.