A total charge of 3.39 C is distributed on two metal spheres. When the spheres a

Question

A total charge of 3.39 C is distributed on two metal spheres. When the spheres are 10.00 cm apart, they each feel a repulsive force of 3.2* 10^11 N. How much charge is on the sphere which has the lower amount of charge? In a right angle triangle ABC, angle ABC is 90 Degree, AB = 2 m, and angle ACB is 41.81 Degree. A point charge of 5* 43 nC is placed at point C, point charge 4* 43 nC is placed at point A and point charge 1C is placed in point B. Calculate the force on charge at C due to others two.

Explanation / Answer

(7). Total charge q + q '= 3.39 C --------( 1)

Separation r = 10 cm = 0.1 m

Force F = 3.2 x10 11 N

We know F = Kqq' /r 2

From this qq' = Fr 2 / K

= (3.2 x10 11 )(0.1) 2/(8.99x10 9 )

= 0.35595

We know (q-q') 2 = (q+q') 2 -4qq'

= 3.39 2 -4(0.35595)

= 10.06

q - q ' = 3.173 ---------( 2)

Eq(1) +eq(2) / 2 ==> q = (3.39+3.173)/2

= 3.281 C

So, q ' = q - 3.173

= 0.1085 C

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