Force of repulsion =F=0.1 N
Distance between two charges =r=5cm=0.05m
Both charges are identical.
Force of same charges when r =2cm=0.02m
F=k×(q1q2)÷r2(whole square)
By using coulombs law we know that;
F=k×(q1q2)÷r2(whole square)
Since;it is provided that q1=q2=q
Hence we can write it as;
F=k(q)2(whole square)÷r2(whole square)
When distance =5cm
Foce F1=k(q)2(whole square)÷(0.05)2(whole square)=0.1 eq 1
When distance is 2cm
=0.02m,F2=?
F2=k(q)2(whole square)÷(0.02)2(whole square) eq 2
Dividing equation 2 by 1
F2÷0.1=(0.05)(0.05)÷(0.02)(0.02)=25÷4
F2=25÷4×0.1=0.62N
F2=0.62N
two charges repel each other with a force of 0.1N when they are 5cm apart. Find the forces between the same charges when they are 2cm and 8cm apart.
3 answers
F= 0.1N
r1= 5 cm
= 0.05m
r2= 4 cm
= 0.04m
F= k(q1 q2)/r2 eq 1
k(q1 q2)= F*r2 put in equation 2
F'=k(q1 q2)/r'2 eq 2
F'=F*r2/r'2
Putting values
F'= 0.1*(0.05)*(0.05)/(0.04)*(0.04)
F'=0.62N
r1= 5 cm
= 0.05m
r2= 4 cm
= 0.04m
F= k(q1 q2)/r2 eq 1
k(q1 q2)= F*r2 put in equation 2
F'=k(q1 q2)/r'2 eq 2
F'=F*r2/r'2
Putting values
F'= 0.1*(0.05)*(0.05)/(0.04)*(0.04)
F'=0.62N
0.62N