# Calculate the charge on an oil droplet

Shaye
Homework Statement:
A tiny negatively charged oil drop is held stationary in the electric field between two horizontal parallel plates, as shown below (please see attachment). Its mass is 4.0x10^-15kg.

Question 15,b) Use the fact that the 2 forces balance to calculate the charge on the oil drop. (g = 10 N Kg^-1).
Relevant Equations:
1. F - Ma
2. E = F/Q or E = V/D (Electric field strength)
The book gives an answer of Q = 3.2 x 10^-19C

I get an answer of 6.67 x 10^-19C. Workings below:  Homework Helper
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The book gives an answer of Q = 3.2 x 10^-19C

I get an answer of 6.67 x 10^-19C.
Hi @Shaye. I agree with your calculation. Looks like a mistake in the book.

Other points:
- don’t forget the minus sign for the charge;
- round the answer to 2 significant figures;
- in your intermediate step, when you find the value of the force, it’s good practice to include the unit (N);
- consider working in symbols and leaving the arithmetic to the end.

• Shaye and PeroK
Homework Helper
Gold Member
2022 Award
- consider working in symbols and leaving the arithmetic to the end.
I agree. At this level, I would expect you to combine two simple formulas before plugging in the numbers:
$$mg = qE = \frac{qV}{d} \ \Rightarrow \ q = \frac{mgd}{V}$$Then you do that on a calculator.

• Shaye and Steve4Physics
Shaye
Hi @Shaye. I agree with your calculation. Looks like a mistake in the book.

Other points:
- don’t forget the minus sign for the charge;
- round the answer to 2 significant figures;
- in your intermediate step, when you find the value of the force, it’s good practice to include the unit (N);
- consider working in symbols and leaving the arithmetic to the end.
Thanks @Steve4Physics

• Steve4Physics
Shaye
I agree. At this level, I would expect you to combine two simple formulas before plugging in the numbers:
$$mg = qE = \frac{qV}{d} \ \Rightarrow \ q = \frac{mgd}{V}$$Then you do that on a calculator.
Yeah I should probably start to do this going forward @PeroK Good tip!