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?

2.2 µC
680 µC
2.2 C
680 C

Answer 1

The electric field produced by a charge is given by:

`E=k(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=(Er^2)/(k)=((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.

Answer 2

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 =
`(1)/(4\pi \epsilon_(0))(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* 4\pi \epsilon_(0)* R^2`

=
`1236* (1)/(9* 10^9)* (4)^2\ Coulomb`

=
`2197.33* 10^(-9)\ C`

=
`2.19733* 10^(-6)\ C`

=
`2.2\ \mu C`

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