Question

82. Two identical metal balls carry charges of
+2 u c and -12 ulc and attract each other by a
force 'F'. The balls are kept in contact and
moved back to their initial positions. The force
between them now is
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Answer 1

Answer:hii

Explanation:

Hii answer is in the attached picture

Answer 2

Answer:

The force between them now is $\frac{-25}{24} F$.

Explanation:

The initial force (F) between the charges is given as,

$F=\frac{k\times 2\mu \times -12\mu}{r^2}$      (1)

k=coulombs constant

r=separation between the charges

Since the balls are kept in contact and moved back to their initial positions there will be equal distribution of charges between the balls i.e.

$\frac{2\mu-12\mu}{2} =-5\mu C$     (2)

Now the final force (F') between them is,

$F'=\frac{k\times -5\mu \times -5\mu}{r^2}$     (3)

We can get the following result by dividing equations (1) and (2):

$\frac{F}{F'}=\frac{2\mu\times -12\mu}{-5\mu\times -5\mu}$

$F'=F\times \frac{-5\mu\times -5\mu}{2\mu\times -12\mu}$

$F'=\frac{-25}{24} F$  (4)

Hence, the force between them now is $\frac{-25}{24} F$.

#SPJ3

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