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Two otherwise identical, small conducting spheres have charges +7.0 µC and −3.0 µC. When placed a distance r apart, each experiences an attractive force of 3.7 N. The spheres are then touched together and moved back to a distance r apart. Find the magnitude of the new force on each sphere.

Respuesta :

Answer:

0.7 N

Explanation:

q1 = 7 micro coulomb

q2 = - 3 micro coulomb

distance = r

Force, F = 3.7 N

Let the new force be F'.

By use of Coulomb's law

[tex]F=\frac{Kq_{1}q_{2}}{r^{2}}[/tex]

[tex]3.7=\frac{K\times 7\times 10^{-6}\times 3\times 10^{-6}}{r^{2}}[/tex]        .... (1)

Now touch the spheres,

q1 = q2 = (7 - 3 )/ 2 = 2 x 10^-6 C

Now

[tex]F'=\frac{K\times 2\times 10^{-6}\times 2\times 10^{-6}}{r^{2}}[/tex]     .... (2)

Dividing equation (2) by equation (1), we get

[tex]\frac{F'}{3.7}=\frac{4}{21}[/tex]

F' = 0.7 N

Thus, the new force is 0.7 N.

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