Question

A particle has a charge of -4.25 nC.

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

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C$

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

$E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m$

The distance from the electric field is 1.71436 m