Answer:
Explanation:
Force of attraction between two charges is 36N
Then from coulombs electrostatic law.
The force of attraction or repulsion between two charges (q1 and q2) is directly proportional to the product of the force and inversely proportional to the square of the distance apart.
F = kq1•q2 / r².
kq1•q2 / r² = 36
Now, we are told that the second charges has increased by a factor of 3
Then,
q2' = 3q2
So, since the first charge is contact, the radius is constant and k is constant,
Then,
F is directly proportional to the second charge, so if the second charge is triple then the force is tripled
Then,
F' = 3F = 3 × 36 = 108 units
Unit of force is newton
F' = 108N
Also,
F' = kq1•q2' / r²
F' = kq1•3q2 / r².
F' = 3 × kq1•q2 / r²
Since, kq1•q2 / r² = 36
Then,
F' = 3 × 36
F' = 108 units
F' = 108N
The new electrostatic force is 108 units.
Let the charges on object 1 and object 2 be q and Q respectively.
The electrostatic force on charge q due to charge Q, separated by a distance r is given by:
F = kqQ/r²
where k is Coulomb's constant
According to the question:
F = 36 units
⇒ kqQ/r² = 36 units
Now, the charge on object 2 is tripled, so it now acquires a 3Q charge.
So the new electrostatic force will be:
F' = kq(3Q)/r²
F' = 3kqQ/r²
F' = 3F = 3×36
F' = 108 units
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