Two charged particles, q1 and q2, are located on the x-axis, with q1 at the origin and q2 initially atx1 = 14.7 mm. In this configuration, q1 exerts a repulsive force of 2.62 µN on q2. Particle q2 is then moved tox2 = 18.0 mm. What is the force (magnitude and direction) that q2 exerts on q1 at this new location? (Give the magnitude in µN.)

To magnitude of the force that exerts on at this new location is 1.75 µN.

Given the following data:

Scientific data:

To calculate the force (magnitude and direction) that exerts on at this new location, we would apply Coulomb's law:

Mathematically, the force in an electric field is given by this formula:

Where:

For the force, when r = 0.018:

Note: 1 µN =

Force, F = 1.75 µN.

In conclusion, the direction of the force at the new location is outward from the origin along .

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The force magnitude is 1.75 μN acting outward from the origin towards x₂

Explanation:

The given parameters, are;

The location of q₁ = The origin

The location of q₂ = 14.7 mm from q₁

The repulsive force exerted on q₂ by q₁ = 2.62 μN

The location the particle q₂ is located to =  18.0 mm from q₁

By Coulomb's law, we have;

Where;

k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)

r = The distance between the particles

F = The force acting between the particles

When r = 14.7 mm F = 2.62 μN

∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸

q₂ × q₁ = 1.48 × 10⁻¹⁸ C²

When the distance is increased to 18.0 mm, we have;

F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N

∴ F = 1.75 μN.

Therefore;

The force magnitude is 1.75 μN outward from the origin towards  x₂.

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27

+15-35=-20

going in opposite directions