Calculate the charge on an oil droplet

In summary, the conversation discusses a calculation involving charge, with the book providing an answer of Q = 3.2 x 10^-19C and the speaker getting an answer of 6.67 x 10^-19C. There is also a mention of rounding, including units in intermediate steps, and using symbols instead of arithmetic.
  • #1
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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:

20220417_110122.LARGE.jpeg
20220407_114835.LARGE.jpeg
 
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  • #2
Shaye said:
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.
 
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  • #3
Steve4Physics said:
- 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.
 
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  • #4
Steve4Physics said:
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
 
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  • #5
PeroK said:
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!
 

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