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Expert-verified Found in: Page 664 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # What is the repulsive force between two pith balls that are 8.00cm apart and have equal charges of -30.0 nC ?

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

See the step by step solution

## Step 1: Given Data

• Separation between pith balls = 8.00 cm
• Charge on each pith ball = 30.0 nC

## Step 2: Electrostatic force

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

## Step 3: 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 $\left(K=9×{10}^{9}N-{m}^{2}/{C}^{2}\right)$ , 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,

Hence, The two pith balls will experience a force of repulsion of $1.26×{10}^{-3}N$ . ### Want to see more solutions like these? 