Question

Sipas spectator (7%) Problem 2: Suppose a speck of dust in an electrostatic precipitator has N = 1.75 x 1017 protons in it and carries a net charge of Q = -37 nC. 4 50% Part (a) Enter an expression for the number of electrons N in the speck of dust in terms of the elementary charge, 90, and variables from the problem statement. Grade Summary Ne = 1 Deductions 0% Potential 100% a HOME B d 0 g ( 7 8 1^ ^ 4 5 / 1 2 + - 0 VO BACKSPACE k 9 6 3 . DEL Submissions Attempts remaining: 3 (3% per attempt) detailed view N P END CLEAR t Submit Feedback I give up! Hints: 1% deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback. HA 50% Part (b) How many electrons are in the speck of dust? >> A 50% Part (b) How many electrons are in the speck of dust? Ne = 1 Grade Summary Deductions 0 % Potential 100% HOME - E sin cos tan cotan asin acos atan) acotan sinh) cosh) tanho cotanh) Degrees Radians ( ) 7 8 1^ ^ 4 5 / 1 2 + - 0 VO BACKSPACE 9 6 3 . DEL Submissions Attempts remaining: 3 (3% per attempt) detailed view END CLEAR Submit Hint Feedback I give up!

Answer 1

Answer: 2.31 x 10^9

Step-by-step explanation:

To find the number of electrons (Ne) in the speck of dust, you can use the relationship between charge and the elementary charge (e):

Ne = Q / e

Given that the charge Q is -37 nC (nanoCoulombs) and the elementary charge e is approximately 1.602 x 10^-19 Coulombs, you can calculate Ne as follows:

Ne = (-37 x 10^-9 C) / (1.602 x 10^-19 C/e)

Ne ≈ -2.31 x 10^9 electrons

So, there are approximately 2.31 x 10^9 electrons in the speck of dust.