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In a vacuum, two particles have charges of q1 and q2, where q1 = +5.1C. They...

In a vacuum, two particles have charges of q1 and q2, where q1 = +5.1C. They are separated by a distance of 0.24 m, and particle 1 experiences an attractive force of 2.8 N. What is the value of q2, with its sign?

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Homework Answers

Answer #1

We need to employ columb's law which is given below:

Here q1=5.1C, r=0.24 m, F=2.8 N, is permittivity of freespace(vaccuum) and is given by 8.854*10-12 F/m. Substituting these values in the above equation, we can solve for q2.

Sign of q2

Since q1 experiencies an attractive force, q1 and q2 are oppositely charged. Since q1 is positively charged, q2 is negative charged.

Therefore, q2=-3.5193 pC

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