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nata0808 [166]
2 years ago
15

compute the electric field experienced by a test a charge q = +0.08µC from a source charge q = +15µC in a vacuum when the test c

harge is placed 0.02 m away from the other charge.
Physics
1 answer:
Fudgin [204]2 years ago
3 0

<u>here</u><u> </u><u>is</u><u> </u><u>your</u><u> </u><u>answer</u><u> </u><u>hope</u><u> </u><u>it's</u><u> </u><u>helpful</u><u> </u><u>for</u><u> </u><u>you</u><u> </u><u>thankyou</u><u> </u><u>I</u><u> </u>

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A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
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Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

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let q1 = - 8.0 x 10^-6 C

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On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
4 years ago
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Answer:

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Explanation:

The time taken can be found by using the formula

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We have the final answer as

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