The new force is 0.015 N
Explanation:
F= [tex]\frac{1}{4\pi(e) } q_{1} (\frac{q_{2} }{r^{2} } )[/tex]
= [tex]k(q_{1} )( \frac{q_{2}}{r^{2} } )[/tex]
Let the initial distance be r
So Force, F = k X [tex]q_{1}[/tex] X [tex]\frac{q_{2} }{r^{2} }[/tex]
0.24 = k X [tex]q_{1}[/tex]X [tex]\frac{q_{2} }{r^{2} }[/tex] - 1
When the distance is increased by factor 4
Then [tex]F_{2}[/tex] = k X [tex]q_{1}[/tex] X [tex]\frac{q_{2} }{r^{2} }[/tex] -2
Dividing equation 1 and 2
[tex]\frac{0.24}{F_{2} }[/tex] = [tex]\frac{16r^{2} }{r^{2} }[/tex]
[tex]\frac{0.24}{F_{2} }[/tex] = 16
[tex]F_{2}[/tex] = 0.015 N
The new force is 0.015 N
