According to Coulomb's law, when the distance between two point charges doubles, what happens to the electric force acting between the charges?

If the distance between charges is doubled, then the Force will become one-fourth.

F' = F/4

Explanation:

The Coulomb's Law gives the relation for the attractive or repulsive forces between two charges. The relation or formula is given as follows:

F = kq₁q₂/r²

where,

F = Attractive or Repulsive forces between the charges

k = Coulomb's Constant

q₁ = magnitude of 1st charge

q₂ = magnitude of 2nd charge

r = distance between charges

If we keep the magnitude of charges constant then the relation between Force an Distance between charges will be:

F α 1/r²

So, it is clear from this relation that if the distance between charges is doubled, then the Force will become one-fourth.

F' = F/4

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j