A cubical volume is bounded by the surfaces $x=0,\ x=a,\ y=0,\ y=a,\ z=0,\ z=a$. The electric field in the region is given by $\vec{E} = E_{0}\,x\,\hat{i}$, where $E_{0} = 4 \times 10^{4}\ \mathrm{N\,C^{-1}\,m^{-1}}$. If $a = 2\ \mathrm{cm}$, the charge contained in the cubical volume is $Q \times 10^{-14}\ \mathrm{C}$. The value of $Q$ is ______. (Take $\epsilon_{0} = 9 \times 10^{-12}\ \mathrm{C^{2}/N\,m^{2}}$)

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Electrostatics

A cubical volume is bounded by the surfaces

x = 0, x = a, y = 0, y = a, z = 0, z = a

. The electric field in the region is given by

E = E_{0} x \hat{i}

, where

E_{0} = 4 \times 1 0^{4} N C^{- 1} m^{- 1}

. If

a = 2 cm

, the charge contained in the cubical volume is

Q \times 1 0^{- 14} C

. The value of

Q

is ______. (Take

ϵ_{0} = 9 \times 1 0^{- 12} C^{2} /N m^{2}

)

Answer: 288

Since the field

E = E_{0} x \hat{i}

points along

x

and increases with

x

, only the two faces perpendicular to the

x

-axis carry any flux. On the face

x = 0

the field is zero, so it contributes nothing; on the face

x = a

the field is

E_{0} a

and the outward flux is

Φ = E_{0} a \cdot a^{2} = E_{0} a^{3}

. By Gauss's law the enclosed charge is

q = ϵ_{0} Φ = ϵ_{0} E_{0} a^{3} = (9 \times 1 0^{- 12}) (4 \times 1 0^{4}) (0.02)^{3}

. With

(0.02)^{3} = 8 \times 1 0^{- 6} m^{3}

, this gives

q = 9 \times 1 0^{- 12} \times 4 \times 1 0^{4} \times 8 \times 1 0^{- 6} = 288 \times 1 0^{- 14} C

. The value of

Q

is therefore

288

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer