+91-9566306857  (or)  +91-9176170648

Ask Questions, Get Answers

Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields

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 $a = 0.1 m.$

1 Answer

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
Download clay6 mobile app