Question

An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between the charges is 30.0 cm and the charge of each is 4.50 pC. (a) Compute the force on the electron at 5.0 cm intervals starting from 5.0 cm from the leftmost charge and ending 5.0 cm from the rightmost charge. (b) Plot the net force versus electron location using your computed values. From the plot, can you make an educated guess as to where the electron feels the least force

Answer 1

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =$k \frac{q_2q_2}{r^2}$

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( $\frac{1}{r_1^2}+ \frac{1}{r_2^2}$)

          F _net = kqQ  ( $\frac{1}{x^2} + \frac{1}{(d-x)^2}$ )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( $\frac{1}{x^2} + \frac{1}{(0.30-x)^2}$ )

          F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m