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A particle has a charge of 1.9x10-6C. What is the charge on a second particle if the electrical force acting on it is 25N at a distance of 0.32m away?

Sagot :

Answer:

1.5x10^-4C

C=Couloumb's

Explanation:

Expression for the electric force between the two charges is given by -

F = (k*q1*q2) / r^2

Here, k = constant = 9 x 10^9 N*m^2 / C^2

F=25N

q1 = 1.9x10^-6 x 10^-6 C

q2 = ?

r = 0.32m

Substitute the given values in the above expression -

25N= 9x10^9 *1.9x10^-6 *q2 / 0.32m^2

25N= 17,100*q2 / 0.1024m

Next part is algebra     multiply both sides by 0.1024 to remove denominator

2.56=17,100*q2  

Divide both sides to isolate q2

q2= 1.5x10^-4C

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