two point charges A and B separated by a distance R attract each other with a force of 12x10^-3N. The force between A and B when the charges on them doubled and distance is halved?
A. 1.92N
B.0.192N
Answers
Answer:
B
0.192
F = KQ1Q2/r^2
Q1 Q2 is double
r is Halfed
Force become 16 times
Given,
When two point charges A and B are separated by a distance R, they attract each other with a force = 12x10^-3N
To find,
The force between A and B when the charges on them doubled and distance is halved.
Solution,
We can simply solve this numerical problem by using the following process:
Mathematically,
The magnitude of the force between two charges Q and q, separated by a distance of R units, respectively, is equal to;
|F| = k × Qq/R^2,
where k = constant of proportionality = 1/(4π×epsilon)
= 8.988 × 10^9 Nm^2/C^2
Now,
The force between A and B in the initial conditions = 12x10^-3 N
=> k × AB/R^2 = 12x10^-3 N
{Equation-1}
Now, according to the question;
changed charges are = 2A and 2B
final distance of separation = R/2
So the force acting between A and B in the final conditions
= k × Qq/R^2
= k × (2A)(2B)/(R/2)^2
= k × (2A)(2B)/(R^2/4)
= 16 × k × AB/R^2
= 16 × 12 × 10^-3 N
(according to equation-1)
= 192 × 10^-3 N
= 0.192 N
Hence, the force between A and B, when the charges on them doubled and distance is halved, is equal to 0.192 N. (Option-B)