Question

Two point charges, the first with a charge of 4.47 x 10-6 C and the second with a charge of 1.86 x 10-6 C, are separated by 17.4 mm. What is the magnitude of the electrostatic force experienced by charge 2

Answer 1

The force between the two charges is 247.15 N.

Step-by-step explanation:

Charge, q = 4.47 x 10^-6 C

charge, q' = 1.86 x 10^-6 C

distance, d = 17.4 mm

Let the force is F.

The force is given by the Coulomb's law:

`F = (K q q')/(r^2)\\\\F =(9* 10^9* 4.47* 10^(-6)*1.86* 10^(-6))/((17.4* 10^(-3))^2)\\\\F = 247.15 N`

Answer 2

Answer:
`247.12\ N`

Step-by-step explanation:

Given

Magnitude of the charges

`q_1=4.47* 10^(-6)\ C`

`q_2=1.86* 10^(-6)\ C`

Distance between them
`d=17.4\ mm`

As both charges are of same sign, they must repel each other

Force experienced by second charge is

`\Rightarrow F_(21)=(kq_1q_2)/(d^2)\\\\\Rightarrow F_(21)=(9* 10^9* 4.47* 10^(-6)* 1.86* 10^(-6))/((17.4* 10^(-3))^2)\\\\\Rightarrow F_(21)=(74.82* 10^(-3))/(302.76* 10^(-6))\\\\\Rightarrow F_(21)=0.2471* 10^3\\\Rightarrow F_(21)=247.12\ N`

Thus, charge 2 experience a force of
`247.12\ N`