Question

A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the mass of the drop is \(1.0 \times 10^{-13} \mathrm{kg}\) and it remains stationary when the potential difference between the plates is $9.76 \mathrm{kV},$ what is the magnitude of the charge on the drop? (Ignore the small buoyant force on the drop.)

Short answer

Answer: The magnitude of the charge on the oil drop is approximately 1.61 x 10^-17 C.

Step-by-step solution

Determine the gravitational force on the oil drop

Using the given mass of the oil drop (\(m = 1.0 \times 10^{-13} \mathrm{kg}\)), we can find the gravitational force acting on it. The gravitational force is given by: \(F_g = m \cdot g\) where \(g\) is the acceleration due to gravity and is approximately equal to \(9.81 \mathrm{m/s^2}\). Therefore, the gravitational force acting on the oil drop is: \(F_g = (1.0 \times 10^{-13} \mathrm{kg})(9.81 \mathrm{m/s^2}) = 9.81 \times 10^{-13} \mathrm{N}\).

Determine the electric field between the plates

We are given that the potential difference between the plates is \(9.76 \mathrm{kV}\), and the distance between the plates is \(16 \mathrm{cm}\). We can convert the distance to meters (\(0.16 \mathrm{m}\)) and the potential difference to volts (\(9760 \mathrm{V}\)). The electric field (\(E\)) between the plates can be calculated using the formula: \(E = \dfrac{V}{d}\) where \(V\) is the potential difference, and \(d\) is the distance between the plates. So, we find: \(E = \dfrac{9760 \mathrm{V}}{0.16 \mathrm{m}} = 61000 \mathrm{V/m}\)

Calculate the electric force on the oil drop

The electric force acting on the oil drop can be calculated using the formula: \(F_e = E \cdot q\) where \(F_e\) is the electric force, \(E\) is the electric field, and \(q\) is the charge on the oil drop. Since the oil drop is stationary, the electric force must be equal to the gravitational force: \(F_e = F_g\) Substituting the equation for the electric force, we get: \(E \cdot q = F_g\)

Solve for the magnitude of the charge on the oil drop

Now that we have found \(E\) and \(F_g\), we can solve for \(q\): \(q = \dfrac{F_g}{E} = \dfrac{9.81 \times 10^{-13} \mathrm{N}}{61000 \mathrm{V/m}} = 1.61 \times 10^{-17} \mathrm{C}\) So, the magnitude of the charge on the oil drop is approximately \(1.61 \times 10^{-17} \mathrm{C}\).