Question

How much electric flux will come out
through a surface S = 10ġ kept in an
electric field Ē =2î +4j+3Â?​

Answer 1

GIVEN :–

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

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

TO FIND :–

$\\ \longrightarrow \: \bf Flux( \phi) =?$

SOLUTION :–

• We know that –

$\\\implies \red{\large{ \boxed{\bf \phi=\overrightarrow{E}.\overrightarrow{ds}}}}$

• Now put the values –

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

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

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

$\\\implies\bf \phi=20(0) +40 (1) +30(0)$

$\\\implies\bf \phi=0+40+0$

$\\\implies \large \red{ \boxed{\bf \phi=40}}$

Answer 2

Answer:

Given :-

• A surface S = $\sf 10\hat{j}$ kept in an electric field $\sf \overrightarrow{E} =\: 2\hat{i} + 4\hat{j} + 3\hat{k}$.

To Find :-

• What is the electric flux come out through a surface.

Formula Used :-

$\bigstar \: \boxed{\sf{Electric\: Flux (\phi) =\: \overrightarrow{E}. \overrightarrow{ds}}}$

where,

• $\sf \phi$ = Electric Flux
• $\sf \overrightarrow{E}$ = Electric Field
• $\sf \overrightarrow{ds}$ = Surface

Solution :-

Given :

• Electric Field = $\sf 2\hat{i} + 4\hat{j} + 3\hat{k}$
• Surface = $\sf 10\hat{j}$

According to the question by using the formula we get,

$\longmapsto \sf Electric\: Flux\: =\: (2\hat{i} + 4\hat{j} + 3\hat{k}) . (10\hat{j})$

$\longmapsto \sf Electric\: Flux\: =\: (2\hat{i}).(10\hat{j}) +\: (4\hat{j}).(10\hat{j}) +\: (3\hat{k}).(10\hat{j})$

$\longmapsto \sf Electric\: Flux\: =\: 2 \times 10(\hat{i}.\hat{j}) +\: 4 \times 10(\hat{j}.\hat{j}) +\: 3 \times 10(\hat{k}.\hat{j})$

$\longmapsto \sf Electric\: Flux\: =\: 20(\hat{i}.\hat{j}) + 40(\hat{j}.\hat{j}) + 30(\hat{k}.\hat{j})$

Now, as we know that,

$\leadsto$ $\sf \hat{i} . \hat{j}$ = 0

$\leadsto$ $\sf \hat{j} . \hat{j}$ = 1

$\leadsto$ $\sf \hat{k} . \hat{j}$ = 0

$\longmapsto \sf Electric\: Flux\: =\: 20(0) + 40(1) + 30(0)$

$\longmapsto \sf Electric\: Flux\: =\: 20 \times 0 + 40 \times 1 + 30 \times 0$

$\longmapsto \sf Electric\: Flux\: =\: 0 + 40 + 0$

$\longmapsto \sf Election\: Flux\: =\: 0 + 0 + 40$

$\longmapsto \sf Electric\: Flux\: =\: 0 + 40$

$\longmapsto \sf\bold{\pink{Electric\: Flux\: =\: 40}}$

$\therefore$ The electric flux will come out through a surface is 40.