Question

a charge Q uC is placed at the centre of a cube . what would be the flux through one face​

Answer 1

Solution :

⏭ Given:

✏ A charge of Q $\mu{C}$ is placed at the centre of a cube.

⏭ To Find:

✏ Electric flux through one face of cube.

⏭ Concept:

✏ As per gauss's law, Electric flux is defined as ratio of total charge covered by the close surface to the permittivity of medium.

⏭ Formula:

✏ Formula of Electric flux in terms of charge and permittivity is given by

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{ \large{ \phi = \dfrac{ \sum{q}}{ \epsilon_{o}} }}}}}} \: \star$

⏭ Terms indication:

✏ $\phi$ denotes flux

✏ $\sum$q denotes total charge covered by close surface

✏ $\epsilon_o$ denotes permittivity of medium

⏭ Calculation:

• Flux linked with total surface

$\leadsto \sf \: \red{ \phi = \dfrac{Q \times {10}^{ - 6} }{ \epsilon_{o}}}$

• Flux linked with one face of cube

✏ Since, cube has six faces, flux linked with one face is given by

$\leadsto \: \underline{ \boxed{ \bold{ \sf{ \orange{ \large{ \phi = \dfrac{Q \times {10}^{ - 6} }{6 \epsilon_{o}}}}}}}} \: \gray{ \bigstar}$

⏭ Additional information:

• Flux is defined as number of electric field lines.
• SI unitof flux is $\sf{\dfrac{Nm^2}{C}}$
• Electric flux is scalar quantity.

Answer 2

Answer:

$\large\boxed{\sf{\dfrac{q\times{10}^{-6}}{6{\epsilon}_{0}}}}$

Explanation:

It's being given that there is a cube.

Also, a charge is placed at the centre of cube.

• Magnitude of charge = q μC

To find the flux linkage through one of its face.

Let the flux be denoted by $\bold{\phi}$.

We know that, flux through any closed surface is given by the formula,

• $\large \boxed{ \red{\phi = \frac{q_{enclosed}}{ {\epsilon}_{0}} }}$

So, for the given cube, we have ,

Total flux through it is equal to,

$= > { \phi}_{Total} = \dfrac{q \times {10}^{ - 6} }{ {\epsilon}_{0}}$

But, this flux is due to all 6 faces of a cube.

Therefore, for one face, we have,

Flux linked with one face will be,

$= > \phi = \dfrac{q \times {10}^{ - 6} }{6 {\epsilon}_{0}}$

Hence, the required flux is $\bold{\dfrac{q \times {10}^{ - 6} }{6 {\epsilon}_{0}}}$