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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?


2.2 µC

680 µC

2.2 C

680 C

Verified Answer :

skyluke89

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.

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 \(2.2\ \mu C\)

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