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

Answer:

q=0.06 C

Explanation:

Given that

Net flux $\phi = 6.78\times 10^{9}\ Nm^2/C$

Lets take charge inside Gaussian surface = q C

We know that

$\phi=\dfrac{q}{\varepsilon _o}$

$\varepsilon _o=8.854\times 10^{-12} farads\ per\ meter$

Now by putting the values in the above equation we get

$6.78\times 10^{9}=\dfrac{q}{8.854\times 10^{-12}}$

$q=6.78\times 10^{9}\times 8.854\times 10^{-12}\ C$

q=0.06 C

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

q=0.06 C