Respuesta :
Answer:
W = 0.018 J
Explanation:
As we know that work done to change the position of charge is given as change in the potential energy of the system
here we can say
[tex]W = U_f - U_i[/tex]
so here we have
[tex]U = \frac{kq_1q_2}{r}[/tex]
so we can say
[tex]W = \frac{kq_1q_2}{r_2} - \frac{kq_1q_2}{r_1}[/tex]
now plug in all values in it
[tex]W = (9\times 10^9)(1\times 10^{-6})(-2\times 10^{-6})(\frac{1}{1} - \frac{1}{0.5})[/tex]
[tex]W = 0.018 J[/tex]
The minimum amount of work needed to move the charges apart to double the distance between them is 0.018 J.
What is the Work done?
We know to change the position of a charge from one point to another work done can be written as,
[tex]W = U_f - U_i[/tex]
Also, we know that the potential energy can be written as,
[tex]U = k\dfrac{q_1q_2}{r}[/tex]
[tex]W = (k\dfrac{q_1q_2}{r})_f - (k\dfrac{q_1q_2}{r})_i[/tex]
[tex]W =(kq_1q_2)[\dfrac{1}{r_f}-\dfrac{1}{r_i}][/tex]
Given to us
q₁ = +1.0 μC = 1 x 10⁻⁶ C
q₂ = -2.0 μC = 2 x 10⁻⁶ C
[tex]r_f[/tex] [tex]=2r_i = 2 \times 0.5 = 1[/tex] m
[tex]r_i[/tex] = 0.5 m
Substitute the values,
[tex]W = (9 \times 10^{9})(1 \times 10^{-6})(-2 \times 10^{-6})[\dfrac{1}{1}-\dfrac{1}{0.5}][/tex]
[tex]W = (9 \times 10^{9})(1 \times 10^{-6})(-2 \times 10^{-6})[-1]\\\\W = 0.018\rm\ J[/tex]
Hence, the minimum amount of work needed to move the charges apart to double the distance between them is 0.018 J.
Learn more about Work:
https://brainly.com/question/3902440
