Question

two point charges 5 micro C and 10 micro C are separated by a distance r in air. If an additional charge of -4micro C is given to each, by what factor does the force between the charges change

Answer 1

q1= 5 x μC

q2 = 10 x  μC

distance = r 

F= k x  5 x 10 x / r 

F=k 50 /r

New charges 

Q1 = -4μC+ q1 

Q2= -4μC+ q2 

F' = K x Q1 X Q2 / r

F' = K 6 / r 

F/ F' = $\frac{ K 50 /r}{ K 6 /r}$

F/ F' = 50 / 6 

F / F' = 25 / 3 

F' = 3 / 25 F

by 3 / 25 

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Answer 2

Explanation:

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