Question

How much electric fluux will come out
through a surface S = 10g kept in an
electric field E =2i+4j+3A?​

Answer 1

Step-by-step explanation:

ϕ=

E

.

ds

=(2

i

^

+4

j

^

+7

k

^

).(10

j

^

)

=4×10

=40units

Answer 2

$\underline{\boxed{\red{\sf Given:- }}}$

$\begin{gathered} \\ \longrightarrow \: \bf Electric \: \: field(\overrightarrow{E}) = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}\\ \end{gathered}$

$\begin{gathered} \\ \longrightarrow \: \bf Surface( \overrightarrow{ds}) = 10\hat{j}\\ \end{gathered}$

$\underline{\boxed{\green{\sf To \: Find:-}}}$

$\begin{gathered} \\ \longrightarrow \: \bf Flux( \phi) =?\\ \end{gathered}$

$\underline{\boxed{\red{\sf Solution:- }}}$

$\underline{\boxed{\pink{\sf we \: know \: that}}}$

$\begin{gathered} \\\implies \orange{\large{ \boxed{\bf \phi=\overrightarrow{E}.\overrightarrow{ds}}}}\\ \end{gathered}$

• • Now put the values –

$\begin{gathered} \\\implies\bf \phi=(2 \hat{i} + 4 \hat{j} + 3 \hat{k}).(10\hat{j})\\ \end{gathered}$

$\begin{gathered} \\\implies\bf \phi=(2 \hat{i}).(10\hat{j}) + (4 \hat{j}).(10\hat{j}) + (3 \hat{k}).(10\hat{j})\\ \end{gathered}$

$\begin{gathered} \\\implies\bf \phi=20(\hat{i}.\hat{j}) +40 (\hat{j}.\hat{j}) +30 ( \hat{k}.\hat{j})\\ \end{gathered}$

$\begin{gathered} \\\implies\bf \phi=20(0) +40 (1) +30(0)\\ \end{gathered}$

$\begin{gathered} \\\implies\bf \phi=0+40+0\\ \end{gathered}$

$\begin{gathered} \\\implies \large \purple{ \boxed{\bf \phi=40}}\\ \end{gathered}$