Two identical spherical balls, P and Q, each of mass 100 g, are suspended at the same point from a ceiling by means of identical light, inextensible insulating strings - NSC Physical Sciences - Question 7 - 2016 - Paper 1

The electrostatic force of attraction between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using the vertical forces, we have:

$T ext{cos} 20^{ ext{o}} = mg$

Substituting in the known values:

Using Coulomb's law:

$F_e = \frac{k \cdot Q_1 \cdot Q_2}{r^2}$

Where:

• $F_e$ is the electrostatic force
• $k$ is Coulomb's constant, approximately $8.99 \times 10^9 \text{ N m}^2/ ext{C}^2$
• $Q_1 = Q_2 = 250 \times 10^{-9} \text{ C}$

From the free-body diagram:

$F_e = T \text{sin} 20^{ ext{o}}$

Setting these equal gives:

$k \cdot (250 \times 10^{-9})^2 / r^2 = T \cdot \text{sin} 20^{ ext{o}} \implies r \approx 0.0397 ext{ m}$