Question

A charge of 4.5 × 10-9 C is located 3.2 m from a charge of -2.8 × 10-9 C. Find the

electrostatic force exerted by one charge on the other.

​

Answer 1

`F = -1.107*10^(-8) N\\|F| = 1.107*10^(-8) N`

Step-by-step explanation:

`q_1 = 4.5 * 10^(-9) C`

`q_2 = -2.8 * 10^(-9) C`

The distance separating the two charges, r = 3.2 m

According to Coulomb's law of electrostatic attraction, the electrostatic force between the two charges can be given by the formula:

`F = (kq_(1) q_(2) )/(r^2)`

Where
`k = 9.0 * 10^9 Nm^2/C^2`

`F = (9*10^9 * 4.5*10^(-9) * (-2.8*10^(-9))/(3.2^2) \\F = (-113.4*10^(-9))/(10.24)\\F = -11.07 *10^(-9)\\F = -1.107*10^(-8)N`

Answer 2

F = 1.1074 ×
`10^(-8)` N

Step-by-step explanation:

An electrostatic force is either a force or attraction or repulsion between two charges. It can be determined by:

F =
`(kq_(1)q_(2) )/(r^(2) )`

where: F is the force, k is a constant,
`q_(1)` is the first charge,
`q_(2)` is the second charge and r the distance between the charges.

Given that: k = 9 ×
`10^(9)` N
`m^(2)`
`C^(-2)`,
`q_(1)` = 4.5 ×
`10^(-9)`C,
`q_(2)` = -2.8 ×
`10^(-9)`C and r = 3.2 m.

Then,

F =
`(9*10^(9)*4.5*10^(-9)*2.8*10^(-9) )/(3.2^(2) )`

=
`(1.134*10^(-7) )/(10.24)`

= 1.1074 ×
`10^(-8)`

The electrostatic force exerted is 1.1074 ×
`10^(-8)` N, and it is a force of attraction.