Question

determine the force of interaction between the charges -30micro coulomb and +20micro coulomb separated by 10 mm apart​

Answer 1

Answer:

Redistribution of charges takes place.

Charge q

1

=3μC

Charge q

2

=8μC

When third charge q

3

=−5μC is added to each, then new charges on q

1

and q

2

will be

q

1

′

=3−5=−2μC

and, q

2

′

=8−5=3μC

Now,

In first case,

40=

4πε

0

1

⋅

r

2

3×8

In second case,

F=

4πε

0

1

×

r

2

(−2×3)

∴

40

F

=

3×8

−2×3

or F=−10N