Question

How strong is the attractive force between a glass rod with a 0.680 µC charge and a silk cloth with a −0.590 µC charge, which are 11.0 cm apart, using the approximation that they act like point charges? (Enter the magnitude in newtons.)

Answer 1

Answer:

$\left |F\right | =0.298\ N$

Explanation:

Data provided:

Charge on the glass rod, Q = 0.680 μC = 0.680 × 10⁻⁶ C

Charge on the silk cloth, q = 0.590 μC = - 0.590 × 10⁻⁶ C

Distance between glass rod and the silk cloth, r = 11.0 cm = 0.11 m

now,

The force between (F) the charges is given as:

$\left |F\right |=\frac{kQq}{r^2}$

where,

k is the Coulomb's constant = 9 × 10⁹ N.m²/C²

on substituting the respective values, we get

$\left |F\right | =\left |\frac{9\times10^9\times0.680\times10^{-6}\times(-0.590\times10^{-6})}{0.11^2}\right |$

or

$\left |F\right | =0.298\ N$