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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]