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iragen [17]
3 years ago
6

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

Physics
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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