Electrostatics
Solver the following exercises
Problem 1
In a mole of 18 grams of water, what is the total negative charge of all the elements?
It is solved in the following way:
1 mole of water
Water molecule 10 electrons
multiply the 10 water molecules so it equals a molecule that is \(6.023X10^{23}\)
\(10X6.023\ X10^{23}\ X\ 1.602X10^{-19}\) \(=964.884.6\)
Problem 2
Two loads are separated by 8m and the magnitude of the force between them is 2N.
For this we use the following formula
\(F1r1^2=\ F2r2^2\)
For that we know the following data.
r1= 8
F1= 2N
F2= ¿?
r2= 4m
We substitute the formula and we solve it
\(\left(2N\right)\left(8m\right)^2\ =\ F_2\ \left(4m\right)^2\)
\(F2\ =\ 2N\ \left(\frac{8m}{4m}\right)^2\)\(=8N\)
Problem 3
How many electrons a and in a charge of \(-38\)\(\mu\)C
The formula we will use will be
\(Ne\frac{Q}{Qe}\)
We hold it
\(\frac{-38X10^{-6}}{1.602X10^{-19}}\ =\ 2.37X10^{14}\)