Question
Explain the term: Electric flux
✓
Answer
$i.$ The number of lines of force per unit area is the intensity of the electric field $\vec{E}$
$\therefore E=\frac{\text { Number of lines of force }}{\text { Area enclosing the lines of force }}$
$\therefore \quad$ Number of lines of force $=(E) \times($ Area $)$
$ii.$ When the area is inclined at an angle $\theta$ with the direction of electric field, the electric flux can be calculated as follows.
Let the angle between electric field $\vec{E}$, and area vector $\overrightarrow{d S}$ be $\theta$, then the electric flux passing through are $dS$ is given by
$d \varnothing=(\text { component of } dS \text { along } \vec{E}) \times(\text { area of } \overrightarrow{d S})$
$d \varnothing= EdS \cos \theta$
$d \varnothing=\vec{E} \cdot \overrightarrow{d S}$
Total flux through the entire surface .
$\varnothing=\int d \varnothing=\int_S \vec{E} \cdot d \vec{S}=\vec{E} \cdot \vec{S}$
$iii.$ The $SI$ unit of electric flux can be calculated using,
$\varnothing=\vec{E} \cdot \vec{S}=( V / m ) m ^2= Vm$
$[$Note: Area vector is a vector whose magnitude is equal to area and is directed normal to its surface$]$
$\therefore E=\frac{\text { Number of lines of force }}{\text { Area enclosing the lines of force }}$
$\therefore \quad$ Number of lines of force $=(E) \times($ Area $)$
$ii.$ When the area is inclined at an angle $\theta$ with the direction of electric field, the electric flux can be calculated as follows.
Let the angle between electric field $\vec{E}$, and area vector $\overrightarrow{d S}$ be $\theta$, then the electric flux passing through are $dS$ is given by
$d \varnothing=(\text { component of } dS \text { along } \vec{E}) \times(\text { area of } \overrightarrow{d S})$
$d \varnothing= EdS \cos \theta$
$d \varnothing=\vec{E} \cdot \overrightarrow{d S}$
Total flux through the entire surface .
$\varnothing=\int d \varnothing=\int_S \vec{E} \cdot d \vec{S}=\vec{E} \cdot \vec{S}$
$iii.$ The $SI$ unit of electric flux can be calculated using,
$\varnothing=\vec{E} \cdot \vec{S}=( V / m ) m ^2= Vm$
$[$Note: Area vector is a vector whose magnitude is equal to area and is directed normal to its surface$]$
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