Question

Two charges of -4.6 C and 8.7 C are separated by a distance of 32 cm. What type of force is there goind to be and of which magnitude?

Answer 1

Respuesta:

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Answer 2

Explicación:

data:

q1 = - 4.6 C

q2 = 8.7 C

d = 32 cm

f = ?

conversion of units to the International System

          32 cm ( 1 m / 100 cm) = 0,32 m

         d = 0,32 m

Between the two charges there is an attraction force

calculating the force of attraction

                formula:

            f = k . q1 . q2 / d²

where k is a constant and has a value = 9x10^9 N.m²/C²

          f = 9x10^9 N.m²/C² . - 4,6 C . 8,7 C / ( 0,32 m)²

          f = 3,51x10^12 N

Respuesta : The magnitude of the attractive force is f = 3,51x10^12 N

I hope it helps you...