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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

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

q1 = unknown

F=25N

q2 = 1.9x10^-6 C

r = 0.32 m

Substitute the given values in the above expression -

25=9x10^9 *(q1) * 1.9 x10^ -6/ (0.32m)^2

25= 17,100* (q1)/ 0.1024

multiply both sides by 0.1024 to get rid of the denominator

2.56=17,100*(q1)

Divide both sides by 17,100 to isolate q1

q1=1.5x10^-4C

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