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Q10PE

Expert-verifiedFound in: Page 664

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.

- Separation between pith balls = 8.00 cm

- Charge on each pith ball = 30.0 nC

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

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 $\left(K=9\times {10}^{9}N-{m}^{2}/{C}^{2}\right)$ , *q _{1}* is the charge of the first pith ball $\left({q}_{1}=-30.0nC\right)$ ,

Substituting all known values,

$F=\left|\frac{\left(\times {10}^{9}N-{m}^{2}/{C}^{2}\right)\times \left(-30.0nC\right)\times \left(-30.0nC\right)}{{\left(8.00cm\right)}^{2}}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{\left(9\times {10}^{9}N-{m}^{2}/{C}^{2}\right)\times \left(-30.0nC\right)\times \left(\frac{{10}^{-9}C}{1nC}\right)\times \left(-30.0nC\right)\times \left(\frac{{10}^{-9}C}{1nC}\right)}{{\left[\left(8.00cm\right)\times \left(\frac{1m}{100cm}\right)\right]}^{2}}\right|\phantom{\rule{0ex}{0ex}}=1.26\times {10}^{-3}N$

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

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