Find out the force of attraction between two charges if -a) the distance between them is made four times
b)the distance between them is made 1/4

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Solution

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$F = \frac{1}{4 π \in} \frac{q_{1} q_{2}}{r^{2}} . . . . 1 w h e n d i s \tan c e i s m a d e f o u r t i m e s r_{1} = 4 r F_{1} = \frac{1}{4 π \in} \frac{q_{1} q_{2}}{(4 r)^{2}} F_{1} = \frac{1}{4 π \in} \frac{q_{1} q_{2}}{16 r^{2}} . . . 2 F_{1} = \frac{F}{16} f o r c e d e c r e a s e s t o 16 t i m e s w h e n d i s \tan c e i s m a d e f o u r t i m e s r_{1} = \frac{r}{4} F_{2} = \frac{1}{4 π \in} \frac{q_{1} q_{2}}{(\frac{r}{4})^{2}} F_{2} = \frac{1}{4 π \in} \frac{16 q_{1} q_{2}}{r^{2}} F_{2} = 16 F f o r c e b e c o m e s 16 t i m e s R e g a r d s$

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