80PTS

1 answer:

6 0

A) The magnitude of the electric field generated by a single point charge is given by
$E=k \frac{q}{r^2}$
where k is the Coulomb's constant, q is the charge and r is the distance at which we calculate the field.
Since we want to know the intensity of the field at distance of r=4.52 m from the charge of q=8.15 C, the intensity of the electric field is
$E=k \frac{q}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{8.15 C}{4.52 m}=3.6 \cdot 10^9 N/C$

b) Direction of the field
The charge that produces the field is negative, and we know that the electric field generated by a negative charge points toward the charge. Therefore, the direction of the electric field is towards the charge.

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What electric force would a stationary 3.8 C charge experience if it were far away from any other charges

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Answer:

The electric force will be 0 N

Explanation:

From the question we are told that

The magnitude of the charge is $q_1 = 3.8 \ C$

Generally from Coulombs law the electric force between two charges is mathematically represented as

$F = \frac{ k * q_1 q_2 }{r^2}$

Here r is the distance of separation between that two charges.

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