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musickatia [10]
2 years ago
12

A negative charge, q1, of 6 µC is 0. 002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the e

lectrical force, Fe, applied by q1 on q2? magnitude: 8 × 101 N direction: south magnitude: 8 × 101 N direction: north magnitude: 4 × 104 N direction: south magnitude: 4 × 104 N direction: north.
Physics
1 answer:
prohojiy [21]2 years ago
7 0

Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

<h3>What is electrical force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is

q₁ is the negative charge = 6 µC=6×10⁻⁶ C

q₂ is the positive charge = 3 µC=3×10⁻⁶ C

r is the distance between the charges=0.002 m

F_E is the electric force =?

The value of electric force will be;

\rm F_E= \frac{Kq_1q_2}{r^2} \\\\ F_E= \frac{9\times 10^9\times 6\times 10^{-6}\times3\times10^{-6}}{(0.002)^2}\\\\ \rm F_E=4.05\times10^4\;N

Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

To learn more about the electrical force refer to the link;

brainly.com/question/1076352

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Answer:

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Answer:

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Explanation:

The acceleration of a body is giving by the rate of change of the body's velocity. It is given by

a = Δv / t        ----------------(i)

Where;

a = acceleration (measured in m/s²)

Δv = change in velocity = final velocity - initial velocity   (measure in m/s)

t = time taken for the change (measured in seconds(s))

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i. initial velocity = 5m/s

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Δv = 0 - 5 = -5m/s

ii. time taken = t = 2s

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a = (-5m/s) / (2s)

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