Explore a diverse range of topics and get answers from knowledgeable individuals on IDNLearn.com. Ask your questions and receive detailed and reliable answers from our experienced and knowledgeable community members.

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


Sagot :

Answer:

[tex]Q = 4.9216 * 10^{-7}C[/tex]

Explanation:

Given

[tex]E = 12596.37 N/C[/tex]

[tex]r = 0.593m[/tex]

Required

Determine the magnitude point charge (Q)

This question will be solved using [tex]the\ magnitude[/tex] of the electric field formula

[tex]E = \frac{kQ}{r^2}[/tex]

Where

[tex]k = 9 * 10^9\ Nm^2 / C^2[/tex]

Make Q the subject in [tex]E = \frac{kQ}{r^2}[/tex]

[tex]E * r^2 = kQ[/tex]

[tex]Q = \frac{E * r^2}{k}[/tex]

Substitute values for E, r and k

[tex]Q = \frac{12596.37 * 0.593^2}{9 * 10^9}[/tex]

[tex]Q = \frac{4429.50}{9 * 10^9}[/tex]

[tex]Q = \frac{492.16}{10^9}[/tex]

[tex]Q = 492.16 * 10^{-9}[/tex]

Express in standard form

[tex]Q = 4.9216 * 10^2 * 10^{-9}[/tex]

[tex]Q = 4.9216 * 10^{2-9}[/tex]

[tex]Q = 4.9216 * 10^{-7}C[/tex]