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If two such generic humans each carried 3.0 C coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 700 N weight?

Question

If two such generic humans each carried 3.0 C coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 700 N weight?

check_circleAnswer
Step 1

Given:

Charge on first human = +3 C

Charge in second human = -3 C

Electric attraction between the two given people = 700 N

Step 2

Using Coulomb’s law for calculating the electric force between given people:

According to coulomb's law, the magnitude of force of attraction
between two charges Qi and Q2 is given by, F -Gi|2 (1)
R2
(Where k 9x 10° N m2/C2 and R is
the separation between the charges)
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According to coulomb's law, the magnitude of force of attraction between two charges Qi and Q2 is given by, F -Gi|2 (1) R2 (Where k 9x 10° N m2/C2 and R is the separation between the charges)

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Step 3

Calculating the separation bet...

Substitute Q+3 C, Q2 -3 C and FE
700 N in equation 1,
So. 700N=9x10° Nm° / C2 x(3C) x(3C)
R2
Or R281x107m
700
Or R 10.75x 10'm = 10.75 km
R 10.75 km
help_outline

Image Transcriptionclose

Substitute Q+3 C, Q2 -3 C and FE 700 N in equation 1, So. 700N=9x10° Nm° / C2 x(3C) x(3C) R2 Or R281x107m 700 Or R 10.75x 10'm = 10.75 km R 10.75 km

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Electric Charges and Fields

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