two point charges a and b, having charges+Q and - Q respectively, are placed at certain distance apart and force acting between them is F. If25%charge of A is transferred to b, then force between the charges becomes
Answers
Answer:
Explanation:
F=Q1XQ2/Sq of distance
if 25% of force transfer to another then
Q1= 0.75Q1
Q2=0.75 Q2 Becasue Q2 when take 25% of positive charge then charge left will be 0.75
So F2=
F2 = 0.75x0.75x F
F2 = 9/16 F
Answer:
Effect on the force between two charges a and b having magnitudes (+Q and –Q) and certain distance between them.
Explanation:
By the law (Coulomb’s law) it is obvious that the electrostatic force acting between the two charges separated by a distance d is given by the equation:-
F = K( q1 X q2)/d^2.
Where F is the force acting between charges q1 and q2 while d is the distance between the two charges and K is the constant.
So, by this equation it is obvious that the force is directly proportional to the product of the two charges.
Now, if 25 percent of the charge is transferred from q1 to q2 (or a to b as in this case), still it will be equivalent to the product between them which should in fact remain unaffected. The force in both cases should before and after transfer of charges should in fact remain the same.
However if the value of one of the charges increases or decreases, then the value of the force acting on them will also increase or decrease by the same proportion.