Question

In a vacuum, two particles have charges of 41 and q2, where q-+5.2HC. They are separated by a distance of 0.27 m, and particle 1 experiences an attractive force of 4.8 N. What is the value of q2, with its sign? Numbe UnitsC the tolerance is +/-2% the toleraTLG T TIF7

Answer 1

Solution:-

Electric force is given by:

E=K Q1 Q2 /R^2

Put values..

4.8=9*10^9*5.2*10^-6*Q2/(0.27)^2

Q2=7.48*10^-6 C

=-7.48 uC (Negative as the force is attractive)

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