A cube is placed inside an electric field, ov | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

As electric field is in $y-$direction so electric flux is only due to top and bottom surface

Bottom surface ${y}=0$

$\Rightarrow {E}=0 \Rightarr

Vedclass · Accepted Answer

$8.3$

Vedclass · Answer

As electric field is in $y-$direction so electric flux is only due to top and bottom surface

Bottom surface ${y}=0$

$\Rightarrow {E}=0 \Rightarrow \phi=0$

Top surface ${y}=0.5\, {m}$

$\Rightarrow {E}=150(0.5)^{2}=\frac{150}{4}$

Now flux $\phi={EA}=\frac{150}{4}(.5)^{2}=\frac{150}{16}$

By Gauss's law $\phi=\frac{Q_{	ext {in }}}{\epsilon_{0}}$

$\frac{150}{16}=\frac{{Q}_{	ext {in }}}{\epsilon_{0}}$

${Q}_{	ext {in }}=\frac{150}{16} 	imes 8.85 	imes 10^{-12}=8.3 	imes 10^{-11} \,{C}$

A cube is placed inside an electric field, $\overrightarrow{{E}}=150\, {y}^{2}\, \hat{{j}}$. The side of the cube is $0.5 \,{m}$ and is placed in the field as shown in the given figure. The charge inside the cube is $.....\times 10^{-11} {C}$

Similar Questions

Each of two large conducting parallel plates has one sided surface area $A$. If one of the plates is given a charge $Q$ whereas the other is neutral, then the electric field at a point in between the plates is given by

Draw electric field lines of simple charge distribution.

A charge $'q'$ is placed at one corner of a cube as shown in figure. The flux of electrostatic field $\overrightarrow{ E }$ through the shaded area is ...... .

A point charge $q$ is placed on the centre of a hemispherical surface as shown in figure.The net flux of electric fietd tnroug the hemi-spherical surface is closest to

Two infinite plane parallel sheets separated by a distance $d$ have equal and opposite uniform charge densities $\sigma $. Electric field at a point between the sheets is