The force of repulsion that two like charges exert on each other is 3.4 N. What will the force be if the distance between the charges is increased to 4 times its original value?

1 Answer
SagarStudy
Jan 7, 2016

Let #q_1# and #q_2# be the charges, #K# be the Coulomb's constant and #r# be the distance between the charges when the force is #3.4N# and let #F# denote this force.

#F=(kq_1q_2)/r^2#

#implies 3.4=(kq_1q_2)/r^2#

Now, every other condition remains same except the distance between the charges. The distance between the charges has been increased #4# times its original value. Let #r'# be then new distance between the charges, then
#r'=4r#
Let the new force between the charges be denoted by #F'#.
#F'=(kq_1q_2)/r_'^2#

#implies F'=(kq_1q_2)/(4r)^2#

#implies F'=(kq_1q_2)/(16r^2)#

#implies F'=(1/16)((kq_1q_2)/r^2)#

Since #3.4=(kq_1q_2)/r^2#

#implies F'=(1/16)(3.4)#

#implies F'=0.2125N#

Hence the force between the two charges becomes #0.2125N#

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