Question

What is the repulsive force between two pith balls that are 8.00cm apart and have equal charges of -30.0 nC ?

Short answer

Two pith balls will experience a force of \({\rm{1}}{\rm{.26 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ N}}\) between them.

Step-by-step solution

Given Data

• Separation between pith balls = 8.00 cm

• Charge on each pith ball = 30.0 nC

Electrostatic force

The two similar charges are separated by some distance, then they experience some repulsive force known as the electrostatic force of repulsion.

Repulsive force between two pith balls

The two pith balls will experience a force of repulsion

$F=\left|\frac{K{q}_1{q}_2}{r^2}\right|$

Here, Q is the electrostatic force constant $\left(K=9\times {10}^{9}N-m^2/C^2\right)$ , q₁ is the charge of the first pith ball $\left(q_1=-30.0nC\right)$ , q₂ is the charge on the second pith ball $\left(q_2=-30.0nC\right)$ , and r is the separation between pith balls $\left(r=8.00cm\right)$.

Substituting all known values,

$$\begin{gathered} F=\left|\frac{\left(\times {10}^{9}N-m^2/C^2\right)\times \left(-30.0nC\right)\times \left(-30.0nC\right)}{{\left(8.00cm\right)}^2}\right| \\ =\left|\frac{\left(9\times {10}^{9}N-m^2/C^2\right)\times \left(-30.0nC\right)\times \left({\displaystyle \frac{{10}^{-9}C}{1nC}}\right)\times \left(-30.0nC\right)\times \left(\frac{{10}^{-9}C}{1nC}\right)}{{\left[\left(8.00cm\right)\times \left({\displaystyle \frac{1m}{100cm}}\right)\right]}^2}\right| \\ =1.26\times {10}^{-3}N \end{gathered}$$

Hence, The two pith balls will experience a force of repulsion of $1.26\times {10}^{-3}N$.