Question

How much charge is enclosed by a Gaussian surface through which the electric flux is 6.78 x 10^9 Nm^2\C.

Answer 1

q=0.06 C

Step-by-step explanation:

Given that

Net flux
`\phi = 6.78* 10^(9)\ Nm^2/C`

Lets take charge inside Gaussian surface = q C

We know that

`\phi=(q)/(\varepsilon _o)`

`\varepsilon _o=8.854* 10^(-12) farads\ per\ meter`

Now by putting the values in the above equation we get

`6.78* 10^(9)=(q)/(8.854* 10^(-12))`

`q=6.78* 10^(9)* 8.854* 10^(-12)\ C`

q=0.06 C

Therefore the net charge inside the surface will be 0.06 C.

q=0.06 C