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3 years ago

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A) What magnitude point charge creates a 12596.37 N/C electric held at a distance of 0.593 m?

1 answer:

Oksana_A [137]3 years ago

8 0

Answer:

$Q = 4.9216 * 10^{-7}C$

Explanation:

Given

$E = 12596.37 N/C$

$r = 0.593m$

Required

Determine the magnitude point charge (Q)

This question will be solved using $the\ magnitude$ of the electric field formula

$E = \frac{kQ}{r^2}$ <em> </em>

Where

$k = 9 * 10^9\ Nm^2 / C^2$

Make Q the subject in $E = \frac{kQ}{r^2}$

$E * r^2 = kQ$

$Q = \frac{E * r^2}{k}$

Substitute values for E, r and k

$Q = \frac{12596.37 * 0.593^2}{9 * 10^9}$

$Q = \frac{4429.50}{9 * 10^9}$

$Q = \frac{492.16}{10^9}$

$Q = 492.16 * 10^{-9}$

Express in standard form

$Q = 4.9216 * 10^2 * 10^{-9}$

$Q = 4.9216 * 10^{2-9}$

$Q = 4.9216 * 10^{-7}C$

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