Test Index

Electrostatics Part 1

Section: Physics

© examsnet.com

Question : 96 of 100

Marks: +1, -0

If the electric flux entering and leaving an enclosed surface respectively is

\phi_1

\phi_2

, the electric charge inside the surface will be

$(\phi_2-\phi_1) \varepsilon_0$
$\frac{\phi_2+\phi_1}{\varepsilon_0}$
$\frac{\phi_2-\phi_1}{\varepsilon_0}$
$(\phi_1+\phi_2) \varepsilon_0$

Solution:

The flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive, by convention. Therefore the net flux leaving the enclosed surface

=\phi_2-\phi_1

\therefore

the change enclosed in the surface by Gauss's law is

q=\varepsilon_0 (\phi_2-\phi_1)

© examsnet.com

Go to Question:

Prev Question

Next Question