Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields
0 votes

The electric field components in Figure are $E_x = \alpha x^{1/2}, E_y = E_z = 0,$ in which $\alpha = 800 N/C m^{1/2}.$ Calculate the flux through the cube.Assume that $a = 0.1 m.$

Can you answer this question?

1 Answer

0 votes
Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and $\Delta S$ is $\pm \large\frac{\pi}{2}$
Therefore, the flux $\phi = E.\Delta S$ is separately zero for each face of the cube except the two shaded ones.
Now the magnitude of the electric field at the left face is
$E_L = \alpha x^{1/2} = \alpha a^{1/2}$ (x = a at the left face).
The magnitude of electric field at the right face is
$E_R = \alpha x^{1/2} = \alpha (2a)^{1/2} $ (x = 2a at the right face).
The corresponding fluxes are
$\qquad= \alpha a ^{5/2}( \sqrt 2-1)$
$\qquad= 800 (0.1)^{5/2}. (\sqrt 2 -1)$
$\qquad= 1.05 \;Nm^2 C^{-1}$
Hence A is the correct answer.
answered Jun 2, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App