Question

If the electric force between two charges is 4.2 × 10-2 N, what would the new force be if the distance between the charges is doubled and the charge on one of the charges is tripled?

Answer 1

The force becomes 0.0315 N.

Step-by-step explanation:

Force, F = 4.2 x 10^-2 N

When the distance is doubled, a charge is tripled, Let the force is F'.

The force between the two charges is

`F = (K qq')/(r^2)\\`

when, q' = 3 q' and r is 2 r so

`F' = (K 3qq')/(4r^2) = (3 F)/(4) = \frac {3* 4.2* 10^(-2)}{4}=0.0315 N`

Answer 2

New force = 0.063 N

Step-by-step explanation:

Given that,

The electric force between two charges is
`4.2* 10^(-2)\ N`

The formula for the electric force is:

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

If the distance between the charges is doubled, r' = 2r and one of the charges is tripled, q₁' = 2q₁, q₂' = 3q₂

Put all the values,

`F'=(kq_1'q_2')/(r'^2)\\\\(F)/(F')=((kq_1q_2)/(r^2))/((kq_1'q_2')/(r'^2))\\\\(F)/(F')=((q_1* q_2)/(r^2))/((2q_1* 3q_2)/((2r)^2))\\\\(F)/(F')=(4)/(6)=(2)/(3)\\\\F'=(3* 4.2* 10^(-2))/(2)\\\\F'=0.063\ N`

So, the new force is 0.063 N.