Question

An electron is 0.18 cm from an object with an electric charge of +3.0×10−6C3.0×10−6C. Determine the magnitude of the electrical force FOonethat the object exerts on the electron.

Answer 1

The magnitude of the force is 1.33 x
`10^(-9)` N

Step-by-step explanation:

In this question there are two charged body involved - the electron and the charged object. To determine the magnitude of the electrical force of the object on the electron, we use the Coulomb's law which states that:

"The force of attraction or attraction (F) between two charged bodies is the directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them."

Mathematically;

F
`\alpha`
`q_(1) q_(2)` /
`r^(2)`

Where
`q_(1) and q_(2)` are the charges of the two bodies.

=> F = k x |
`q_(1) q_(2)` |/
`r^(2)` -----------------(i)

Where k is a constant and has a value of k = 8.9876 ×
`10^(9)`N
`m^(2) c^(-2)`

Given;

`q_(1)` is the charge of an electron = 1.6 ×
`10^(-19)` C

`q_(2)` is the charge on the object = +3.0 x
`10^(-6)`C

r = 0.18cm = 0.0018m

Substituting the values of k, r,
`q_(1)` and
`q_(2)` into the equation (i) above, we have;

=> F = 8.9876 ×
`10^(9)` x 1.6 ×
`10^(-19)` x 3.0 x
`10^(-6)` ÷
`0.0018^(2)`

=> F = 1.33 x
`10^(-9)` N

Answer 2

F = 1.3*10⁻⁹ N

Step-by-step explanation:

If we can approximate both charged objects as point charges, the force that one object exerts on the other, in magnitude, is given by Coulomb's Law, as follows:

`F=(k*q1*q2)/(r^(2)) (1)`

where k = 9.0*10⁹ N*m²/C², q₁ = e = 1.6*10⁻¹⁹ C, q₂ = +3.0*10⁻⁶ C and r=0.0018 m.

Replacing all these values in (1), we can solve for F as follows:

`F=((9e9 N*m2/C2)*1.6e-19C*3.0e-6C)/((0.0018m)^(2)) = 1.3e-9 N`

⇒ F = 1.3*10⁻⁹ N