Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

What is the magnitude of a point charge that would create an electric ﬁeld of 1.00 N/C at points 1.00m away?

1 Answer

Feb 20, 2016

$| q | = E \frac{r^{2}}{k} = \frac{1 \frac{N}{C} \cdot 1 m^{2}}{8.99 \times 109 N \cdot \frac{m^{2}}{C^{2}}} = 1 .11 \times 10^{- 10} C$ $|q| = Er^2/ k = (1 N/C*1 m^2)/(8.99×109 N·m^2/C^2) =1 .11×10^(−10) C$

Explanation:

The magnitude of the E ﬁeld due to a point charge q at a distance r is given by $E = k \frac{| q |}{r^{2}}$ $E = k|q|/ r^2$ , Here we are given $E and r$ $E " and " r$ , so we can solve for the required charge, $q$ $q$ :
$| q | = E \frac{r^{2}}{k} = \frac{1 \frac{N}{C} \cdot 1 m^{2}}{8.99 \times 109 N \cdot \frac{m^{2}}{C^{2}}} = 1 .11 \times 10^{- 10} C$ $|q| = Er^2/ k = (1 N/C*1 m^2)/(8.99×109 N·m^2/C^2) =1 .11×10^(−10) C$

Related questions

Impact of this question

26387 views around the world

You can reuse this answer
Creative Commons License