Question

A particle has a charge of -4.25 nC.

Part A

Find the magnitude of the electric field due to this particle at a point 0.250 m directly above it.

Part B

Find the direction of the field

up, away from the particle

down, toward the particle

Part C

At what distance from this particle does its electric field have magnitude of 13.0 N/C?

Answer 1

Answer:

Explanation:

q = - 4.25 nC = - 4.5 x 10^-9 C

(A) d = 0.250 m

The formula for the electric field is given by

$E = \frac{1}{4\pi \epsilon _{0}}\frac{q}{d^{2}}$

By substituting the values

$E = \frac{9\times 10^{9}\times 4.5\times10^{-9}}{0.25\times 0.25}$

E = 648 N/C

(B) As the charge is negative in nature so the direction of electric field is towards the charge and downwards.

Answer 2

(a) The magnitude of the electric field is 612 N/C.

(b) The direction of the electric field will be up, away from the particle.

(c) The distance from the particle is 1.71 m.

Magnitude of the electric field

The magnitude of the electric field is calculated as follows;

E = (kq)/r²

where;

k is Coulomb's constant

q is the charge

r is distance

E = ( 9 x 10⁹ x 4.25 x 10⁻⁹)/(0.25 x 0.25)

E =  612 N/C

Direction of the field

The direction of the electric field is always opposite to the direction of the negative charge.

Thus, the direction of the electric field will be up, away from the particle.

Distance from the particle

The distance from the particle is determined using the following formula;

E = (kq)/r²

r² = kq/E

r² = (9 x 10⁹ x 4.25 x 10⁻⁹) / 13

r² = 2.94

r = √2.94

r = 1.71 m

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