Question

What is the magnitude of the force a 25 charge exerts on a 3 mc charge 35cm away?

Answer 1

The magnitude of force between two electric charges can be calculated using Coulomb's Law, which involves the product of the charges and the inverse square of the distance between them.

Step-by-step explanation:

To determine the magnitude of the force exerted by one charge upon another, we can use Coulomb's Law, which states that the force (F) between two charges (q1 and q2) separated by a distance (r) is given by the equation F = k * |q1*q2| / r^2, where k is Coulomb's constant (8.988 × 10^9 N m^2/C^2).

In the case of a 25 microcoulomb (µC) charge exerting a force on a 3 microcoulomb (µC) charge 35 cm away, we first need to convert the charges into coulombs (C) by multiplying them by 10^-6 and the distance into meters (m) by multiplying it by 10^-2.

Therefore, q1 = 25 × 10^-6 C, q2 = 3 × 10^-6 C, and r = 35 × 10^-2 m. Substituting these values into Coulomb's Law equation gives us the magnitude of the force between the two charges.

Answer 2

The electrostatic force between two charges is given by

`F=k_e (q_1 q_2)/(r^2)`
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the charges

In our problem,

`q_1 = 25 \mu C=25 \cdot 10^(-6) C`

`q_2 = 3 mC = 3 \cdot 10^(-3) C`

`r=35 cm=0.35 m`
Therefore the electrostatic force is

`F=(8.99 \cdot 10^9 Nm^2C^(-2)) ((25 \cdot 10^(-6)C)(3 \cdot 10^(-3)C))/((0.35 m)^2)= 5510 N`