Question

The electric field due to charge Q at a distance 10m from it is 900N/C. The magnitude of charge Q is $\left[ {\dfrac{1}{{4\pi {\varepsilon _o}}} = 9 \times {{10}^9}\dfrac{{N{m^2}}}{{{C^2}}}} \right]$
1) ${10^{ - 2}}C$
2) \[{10^{ - 4}}C\]
3) \[{10^{ - 5}}C\]
4) \[{10^{ - 6}}C\]

Answer 1

Hint:Electric field is defined as the electric force per unit charge and the direction of the electric field is taken as the direction of the force that is applied on the positive test charge. The electric field is proportional to the charge and inversely proportional to the distance. The electric field will continue to decrease as a charge is moved further away from the field.

Complete step by step Solution:
Find the magnitude of the charge Q:
$E = \dfrac{{KQ}}{r}$ ;
Here:
E = Electric Field
K = Constant
Q = Charge
r = Distance
We have been given all the values put in the above equation:
$E = \dfrac{{KQ}}{r}$
Write the above equation in terms of Q:
$E = \dfrac{{KQ}}{r}$
\[ \Rightarrow \dfrac{{E \cdot r}}{K} = Q\]
Put in the given values:
\[ \Rightarrow \dfrac{{900 \times 10}}{{9 \times {{10}^9}}} = Q\]
\[ \Rightarrow \dfrac{{100 \times 10}}{{{{10}^9}}} = Q\]
The magnitude of charge is:
\[ \Rightarrow {10^{ - 6}}C = Q\]

Final Answer:Option “4” is correct. Therefore, the magnitude of charge Q is \[{10^{ - 6}}C\].

Note:Here we have been asked about the magnitude of the charge. All the needed values are given to us. Apply the formula of electric field in relation with charge and distance. There has been a long battle between scientists whether light is particle or wave like, light is a form of EM radiation. On a similar basis electric field is considered to be wave like but according to quantum mechanics electric field can be quantized that means electric field just like light consists of discrete parcels, photons, etc.