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

Answer:

0.00712 m

Explanation:

Given:

Charge on first particle (q₁) = 75 nC = $75\times 10^{-9}\ C$

Charge on second particle (q₂) = 75 nC = $75\times 10^{-9}\ C$

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:

$F=\dfrac{k|q_1||q_2|}{d^2}$

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

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

$1.00=\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{d^2}\\\\d^2=\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{1.00}\\\\d=\sqrt{\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{1.00}}\\\\d=0.00712\ m$

Therefore, the distance between the charges is 0.00712 m.