Question

Charge q1 is placed a distance r0 from charge q2 . What happens to the magnitude of the force on q1 due to q2 if the distance between them is reduced to r0/4 ?

What is the electrostatic force between and electron and a proton separated by 0.1 mm?

Answer 1

When the distance between two charges is reduced, the magnitude of the force on one charge due to the other increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.

Step-by-step explanation:

When the distance between two charges, q1 and q2, is reduced from r0 to r0/4, the magnitude of the force on q1 due to q2 increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.

For example, if we have charges q1 = 2 C and q2 = 4 C, and the original distance r0 = 2 m, the force between them would be F = k × (q1 × q2) / r0^2 = k × (2 × 4) / (2^2) = k × 4, where k is the Coulomb's constant.

If we reduce the distance to r0/4 = 0.5 m, the force would become F' = k × q1 × q2) / (r0/4)^2 = k × 4) / (0.5^2) = k ×64, which is 16 times greater than the original force.

Answer 2

The electrostatic force between and electron and a proton is
`F=2.30* 10^(-20)\ N`

Step-by-step explanation:

It is given that, charge
`q_1` is placed at a distance
`r_o` from charge
`q_2`. The force acting between charges is given by :

`F=(kq_1q_2)/(r_o^2)`

We need to find the force if the distance between them is reduced to
`r_o/4`. It is given by :

`F'=(kq_1q_2)/((r_o/4)^2)`

`F'=16* (kq_1q_2)/(r_o^2)`

`F'=16* F`

So, if the the distance between them is reduced to
`r_o/4`, the new force becomes 16 times of the previous force.

The electrostatic force between and electron and a proton separated by 0.1 mm or
`10^(-4)\ m` is :

`F=(kq_1q_2)/(r_o^2)`

`F=(9* 10^9* (1.6* 10^(-19))^2)/((10^(-4))^2)`

`F=2.30* 10^(-20)\ N`

So, the electrostatic force between and electron and a proton is
`F=2.30* 10^(-20)\ N`. Hence, this is the required solution.