"A droplet of oil carries an extra charge of one electron and is held motionless between parallel plates separated by 1.80 cm across which there is a potential difference of 55.5 kV. What is the mass of the droplet?"
I assumed that this whole thing was actually a vertically placed capacitor.
Here is my approach on this:
A free body diagram shows that:
Fg = Fe
mg = qE
mg = qV/d
Thus m = qV/(dg)
Plugging in values : m = (1.6 * 10-19 * 55500)/(0.018 * 9.8)
I find m = 5.04e-14 kg. This is incorrect. Where did I go wrong?
Looks right to me.
You made a small mistake in your calculation. The force equation should be written as:
Fe = Fg
qE = mg
And when rearranged,
m = qE/g
Using the given values: m = (1.6 * 10^-19 C * 55,500 V) / 9.8 m/s^2
Calculating this, m = 8.98 x 10^-20 kg
So the correct mass of the droplet is approximately 8.98 x 10^-20 kg.
Your approach is correct, but there seems to be a small mistake in the calculations.
Let's go through the calculations step by step to find the error:
Given information:
Charge of the droplet: q = 1 electron = 1.6 × 10^(-19) C
Potential difference: V = 55.5 kV = 55.5 × 10^3 V
Distance between plates: d = 1.80 cm = 0.018 m
Acceleration due to gravity: g = 9.8 m/s^2
Applying the equation mg = qV/d, we find:
m = (qV)/(d * g)
= (1.6 × 10^(-19) C * 55.5 × 10^3 V) / (0.018 m * 9.8 m/s^2)
= (1.6 × 55.5 × 10^(-19+3)) / (0.018 * 9.8)
= 88.8 × 10^(-16) / 0.176
= 5.04 × 10^(-14) kg
So, your answer of 5.04e-14 kg is actually correct. There might have been a rounding error or some other mistake while evaluating the expression. Please check your calculations again, and you should get the correct result.