The flux of the electric field vecE = 24 hati + 30 hatj + 28 hatk N C-1 through an area of 20 m2 on the yz plane is

Question

The flux of the electric field vecE = 24 hati + 30 hatj + 28 hatk N C-1 through an area of 20 m2 on the yz plane is (A) 480 N m2 C-1 (B) 600 N m2 C-1 (C) 560 N m2 C-1 (D) 1640 N m2 C-1

Tardigrade Admin · Accepted Answer

Here, E =24 hat i +30 hat j +28 hat k N C -1 S =20 hat i m 2 Electric flux,φ= vecE ⋅ vecS =(24 hat i +30 hat j +28 hat k N C -1) ⋅(20 hat i m 2) =480 N m 2 C -1

Q. The flux of the electric field $E = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} N C^{- 1}$ through an area of 20 $m^{2}$ on the $yz$ plane is

$480 N m^{2} C^{- 1}$

$600 N m^{2} C^{- 1}$

$560 N m^{2} C^{- 1}$

$1640 N m^{2} C^{- 1}$

Here, $E = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} N C^{- 1}$
$S = 20 \hat{i} m^{2}$
Electric flux, $ϕ = E \cdot S$
$= (24 \hat{i} + 30 \hat{j} + 28 \hat{k} N C^{- 1}) \cdot (20 \hat{i} m^{2})$
$= 480 N m^{2} C^{- 1}$

Q. The flux of the electric field E=24i^+30j^​+28k^NC−1 through an area of 20 m2 on the yz plane is

480Nm2C−1

600Nm2C−1

560Nm2C−1

1640Nm2C−1