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 vecE=E0 x hat t . Where E0=4 × 104 NC -1 m -1. If a=2 cm, the charge contained in the cubical volume is Q × 10-14 C. The value of Q is Take .E0=9 × 10-12 C 2 / Nm 2)

Question

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 vecE=E0 x hat t . Where E0=4 × 104 NC -1 m -1. If a=2 cm, the charge contained in the cubical volume is Q × 10-14 C. The value of Q is Take .E0=9 × 10-12 C 2 / Nm 2) (A)  (B)  (C)  (D)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/653878eed7ae9186b32c60d0834e53ff-.png" /> vec E = E 0 × hat i φ text net =φ ABCD = E 0 a ⋅ a 2 ( q text en /ϵ0)= E 0 a 3 q en = E 0 ∈0 a 3 =4 × 104 × 9 × 10-12 × 8 × 10-6 =288 × 10-14 C Q =288

Q. 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=E0​xt^. Where E0​=4×104NC−1m−1. If a=2cm, the charge contained in the cubical volume is Q×10−14C. The value of Q is ___ Take E0​=9×10−12C2/Nm2)

E=E0​×i^ ϕnet ​=ϕABCD​=E0​a⋅a2 ϵ0​qen ​​=E0​a3 qen​=E0​∈0​a3 =4×104×9×10−12×8×10−6 =288×10−14C Q=288

$E = E_{0} \times \hat{i}$
$ϕ_{net} = ϕ_{A BC D} = E_{0} a \cdot a^{2}$
$\frac{q _{en}}{ϵ _{0}} = E_{0} a^{3}$
$q_{e n} = E_{0} \in_{0} a^{3}$
$= 4 \times 1 0^{4} \times 9 \times 1 0^{- 12} \times 8 \times 1 0^{- 6}$
$= 288 \times 1 0^{- 14} C$
$Q = 288$