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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

Respuesta :

skyluke89

The electric field produced by a charge is given by:

[tex] E=k\frac{q}{r^2} [/tex]

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 [tex] E=1236 N/C [/tex] and the distance is r=4 m, so we can rearrange the previous equation to find the magnitude of the charge:

[tex] 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 [/tex]


so, the correct answer is 2.2 µC.

The correct answer to the question is-  [tex]2.2\ \mu C[/tex]

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 = [tex]\frac{1}{4\pi \epsilon_{0}}\frac{Q}{R^2}[/tex]

Here, [tex]\epsilon_{0}[/tex] is called the absolute permittivity of the free space.

Hence, the charge of source is calculated as -

                                         Q = [tex]E\times 4\pi \epsilon_{0}\times R^2[/tex]

                                            = [tex]1236\times \frac{1}{9\times 10^9}\times (4)^2\ Coulomb[/tex]

                                            = [tex]2197.33\times 10^{-9}\ C[/tex]

                                             = [tex]2.19733\times 10^{-6}\ C[/tex]

                                             = [tex]2.2\ \mu C[/tex]

Hence, the charge of source is [tex]2.2\ \mu C[/tex]

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