Question

A 4.0 cm × 4.2 cm rectangle lies in the xy-plane. You may want to review (Pages 664 - 668) . Part A What is the electric flux through the rectangle if E⃗ =(150ı^−200k^)N/C? Φe = N⋅m2/C Previous AnswersRequest Answer Incorrect; Try Again; 5 attempts remaining Part BPart complete What is the electric flux through the rectangle if E⃗ =(150ı^−200ȷ^)N/C? Φe = 0 N⋅m2/C

Answer 1

Answer:

(A). The flux is 0.336 N.m²/C

(B). The flux is zero.

Explanation:

Given that,

Length = 4.2 cm

Width = 4.0 cm

Electric field $E=(150 i-200 k)\ N/C$

Area vector is perpendicular to xy plane

(A). We need to calculate the flux

Using formula of flux

$\phi=E\cdot A$

Where, E = electric field

A = area

Put the value into the formula

$\phi=(150 i-200 k)\times(4.2\times10^{-2}\times4.0\times10^{-2})k$

$\phi=-200\times4.2\times10^{-2}\times4.0\times10^{-2}$

$\phi=-0.336\ N.m^2/C$

(B). Given electric field

$E=(150i-200j)\ N/C$

We need to calculate the flux

Using formula of flux

$\phi=E\cdot A$

Put the value into the formula

$\phi=(150 i-200 j)\times(4.2\times10^{-2}\times4.0\times10^{-2})k$

Here, The component of k is not given

So, the flux is

$\phi=0$

Hence, (A). The flux is -0.336 N.m²/C

(B). The flux is zero.

Answer 2

Answer:

Explanation:

Area, A = 4 cm x 4.2 cm = 16.8 cm²

A).

$\overrightarrow{E}=150\widehat{i}-200\widehat{k}$

Area is in x y plane so

$\overrightarrow{A}=16.8\times 10^{-4}\widehat{k}$

Electric flux,

$\phi =\overrightarrow{E}.\overrightarrow{A}$

$\phi =\left ( 150\widehat{i}-200\widehat{k} \right ).\left (16.8\times 10^{-4}\widehat{k} \right )$

Ф = 0.336 Nm²/C

B).

$\overrightarrow{E}=150\widehat{i}-200\widehat{j}$

$\phi =\overrightarrow{E}.\overrightarrow{A}$

$\phi =\left ( 150\widehat{i}-200\widehat{j} \right ).\left (16.8\times 10^{-4}\widehat{k} \right )$

Ф = 0 Nm²/C