Question

A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?

Answer 1

The correct answer to the question is-  $2.2\ \mu C$

CALCULATION:

As per the question, the electric field generated by the source charge is 1236  N/C at a distance of 4 m.

Hence , electric field  E =  1236 N/C.

The distance of the point R = 4m

We are asked to calculate the charge possessed by the source.

The electric field produced by a source charge of Q at a distance R is calculated as -

                    Electric field E = $\frac{1}{4\pi \epsilon_{0}}\frac{Q}{R^2}$

Here, $\epsilon_{0}$ is called the absolute permittivity of the free space.

Hence, the charge of source is calculated as -

                                         Q = $E\times 4\pi \epsilon_{0}\times R^2$

                                            = $1236\times \frac{1}{9\times 10^9}\times (4)^2\ Coulomb$

                                            = $2197.33\times 10^{-9}\ C$

                                             = $2.19733\times 10^{-6}\ C$

                                             = $2.2\ \mu C$

Hence, the charge of source is $2.2\ \mu C$

Answer 2

The electric field produced by a charge is given by:

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge that produces the field

r is the distance from the charge

In this problem, the magnitude of the electric field is $E=1236 N/C$ and the distance is r=4 m, so we can rearrange the previous equation to find the magnitude of the charge:

$q=\frac{Er^2}{k}=\frac{(1236 N/C)(4 m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}}=2.2 \cdot 10^{-6} C=2.2 \mu C$

so, the correct answer is 2.2 µC.