Question

Find the magnitude of a charge which produces an electric field of strength is $ 18 \times {10^3}N.{C^{ - 1}} $ at a distance of 5 m in air.
(A) $ 50\mu C $
(B) $ 100\mu C $
(C) $ 25\mu C $
(D) $ 250\mu C $

Answer 1

Hint: As an electric field is given and also the distance between each charge is given. So, we will apply the formula of an electric field in terms of its charge and the distance, i.e. $ \therefore E = \dfrac{1}{{4\pi {\varepsilon _0}}}.\dfrac{q}{{{r^2}}} $ , in which we have already the value of an electric field. So, we can easily calculate the value of electric charge on it.

Complete Step By Step Answer:
Given strength of an electric field is, $ E = 18 \times {10^3}N.{C^{ - 1}} $
Distance, $ r = 5cm $
Now, we will use the formula of electric field in terms of its charge and the distance:
 $ \therefore E = \dfrac{1}{{4\pi {\varepsilon _0}}}.\dfrac{q}{{{r^2}}} $ ……(i)
where, $ E $ is the electric field strength,
 $ q $ is the charge on an electric field,
 $ r $ is the distance between them.
 $ \because \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9} $
So, in eq(i)-
 $
   \Rightarrow 18 \times {10^3} = 9 \times {10^9}\dfrac{q}{{25}} \\
   \Rightarrow q = 50 \times {10^{ - 6}}C \\
  $
We can also write as:
 $ \therefore q = 50\mu C $
Therefore, the required charge in an electric field is $ 50\mu C $ .
Hence, the correct option is (A) $ 50\mu C $ .

Note:
 As we know, the strength of an electric field is also known as electric field intensity. It's a metric for determining the strength of an electric field at any given location. The ratio of the proton's weight to the charge determines the intensity of the electric field, and electric field intensity is a vector quantity.