Expression for an electric field is given by $\overrightarrow{\mathrm{E}}=4000 \mathrm{x}^{2} \hat{\mathrm{i}} \frac{\mathrm{V}}{\mathrm{m}}$. The electric flux through the cube of side 20 cm when placed in electric field (as shown in the figure) is $\_\_\_\_$ V cm.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Electrostatics

Expression for an electric field is given by

E = 4000 x^{2} \hat{i} \frac{V}{m}

. The electric flux through the cube of side 20 cm when placed in electric field (as shown in the figure) is

____

V cm.

Answer: 640

The field

E = 4000 x^{2} \hat{i}

points along

x

and varies only with

x

, so flux crosses only the two faces perpendicular to the

x

-axis. The cube has one such face at

x = 0

, where

E = 0

and no flux enters, and the opposite face at

x = 0.2

m, where

E = 4000 (0.2)^{2} = 160

V m

^{- 1}

. The outgoing flux through that face is

E \times A = 160 \times (0.2)^{2} = 6.4

V\,m, which in the required units is

6.4 \times 100 = 640

V\,cm.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer