two point charge +3microC and +8microC repel each other with a force of 40N.if a charge of -5microC is added to each of them,then the force between them will become?
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Answered by
82
according to Coloumb's law
F= KQq/r^2
r^2 =KQq/F
given,
F= 40 N
Q=+8 uC
q=+3 uC
hence ,
r^2 = K (8)(3)/40= 3K/5 ----------(1)
again ,
when -5 uC added with each of them
Q'= +8-5= +3 uC
q"=+3 -5 =-2 uC
then ,
F' =KQ'.q'/r^2
=K (3)(2)/r^2
put equation (1)
F' =6K/(3K/5)=10N
hence,
force becomes 10N
F= KQq/r^2
r^2 =KQq/F
given,
F= 40 N
Q=+8 uC
q=+3 uC
hence ,
r^2 = K (8)(3)/40= 3K/5 ----------(1)
again ,
when -5 uC added with each of them
Q'= +8-5= +3 uC
q"=+3 -5 =-2 uC
then ,
F' =KQ'.q'/r^2
=K (3)(2)/r^2
put equation (1)
F' =6K/(3K/5)=10N
hence,
force becomes 10N
Answered by
2
Explanation:
in that formula F= KQ.q/r2
i.e K（3）（-2）/r^2
Now i.e -6K/3K/5
-10N
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