Question

Determine a point charge that can given an electric field of magnitudes 100Nc-1 at a point 50cm away in free space

Answer 1

Answer:

Point charge of 2.78 x 10⁻⁹ C.

Explanation:

E = k q/r²

 Here, E = 100 N/C

           r = 50/100 = 0.5 m

Therefore,

⇒ 100 = (9 x 10⁹)q/(0.5)²

⇒ 100 x (0.25)/(9 x 10⁹) = q

⇒ 2.78 x 10⁻⁹ = q

Answer 2

Given :-

point = 50 cm

Solution :-

E = k q/r²

$\sf \implies 100 = \dfrac{9 \times 10^{9}q }{0.5^{2} }$

$\sf \implies q = \dfrac{100 \times 0.5}{9 \times 10^{9} }$

q = 50/10⁻⁹

q = 2.78 x 10⁻⁹