Question

. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them?

Answer 1

0.00712 m

Step-by-step explanation:

Given:

Charge on first particle (q₁) = 75 nC =
`75* 10^(-9)\ C`

Charge on second particle (q₂) = 75 nC =
`75* 10^(-9)\ C`

Force (F) = 1.00 N

Separation (d) = ?

The magnitude of force is given by Coulomb's law which states that, the magnitude of force acting between two charged particles separated by a distance is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Therefore, the magnitude of force is given as:

`F=(k|q_1||q_2|)/(d^2)`

Where,
`k=9* 10^9\ N\cdot m^2/ C^2` is the coulomb's constant.

Plug in the given values and solve for 'd'. This gives,

`1.00=(9* 10^9* 75.0* 10^(-9)* 75.0* 10^(-9) )/(d^2)\\\\d^2=(9* 10^9* 75.0* 10^(-9)* 75.0* 10^(-9) )/(1.00)\\\\d=\sqrt{(9* 10^9* 75.0* 10^(-9)* 75.0* 10^(-9) )/(1.00)}\\\\d=0.00712\ m`

Therefore, the distance between the charges is 0.00712 m.