Question

Given two positively charged particles, of equal magnitude, separated by a distance, "d". What will happen to the force field between the two particles when "d" is halved? A) remain constant B) decrease by a factor of two C) increase by a factor of four D) nothing; no force field between particles

Answer 1

Increase by a factor of four

Step-by-step explanation:

The force between the two charged particle is directly proportional to the product of charges and inversely proportional to the square of the distance between them. This law is known as Coulomb's law. Mathematically, it can be written as :

`F\propto (Qq)/(d^2)`

`F=k(Qq)/(d^2)`

Where
`k=(1)/(4\pi \epsilon_0)`

And this force divided by electric charge is equal to electric field i.e.

`E=k(Q)/(d^2)`

So, the electric field is inversely proportional to the distance.

`E'=k(Q)/(d'^2)`

`E'=k(Q)/((d/2)^2)`

E' = 4 E

When d is halved, the force field between the particles increase by a factor of four. Hence, the correct option is (c).

Answer 2

Answer;

C) increase by a factor of four

Explanation;

• The size of the force varies inversely as the square of the distance between the two charged particles.
• Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
• For example, If the charges come 10 times closer, the size of the force increases by a factor of 100.
• Additionally, the size of the force is proportional to the value of each charge.