Question

The electric field due to a charge at a distance of √5 m from it is 9000 NC-1 . The magnitude of the charge is
2.5 µC

2.0 µC

1.0 µC

5.0 µC

Answer 1

Answer:

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Answer 2

Answer:

The electric field due to a charge is 9000 NC-1 at a distance of √5 m from it  The magnitude of the charge is 5.0 µC

Explanation:

• The electric field due to a charge (q) at a distance (r) from it is given by the formula

      $E=K\frac{q}{r^{2} }$

      where the constant $K=9*10^{9}$

From the question, we have

the electric field is $E=9000N/C$

the distance at which field is required $r=\sqrt{5} m =2.2m$

rearrange the equation to get the charge $q=\frac{Er^{2} }{K}$

the electric charge is

⇒$q =\frac{9000*(2.2)^{2} }{9*10^{9} }$

⇒$q=4.84*10^{-6} C$

⇒$q$≈ 5.0 µC