Question

Charge of 6 microcoulomb placed inside a box calculate electric flux linked with the box?

Answer 1

Given:

Charge of 6 microcoulomb placed inside a box.

To find:

Electric Flux linked with the box.

Calculation:

According to Gauss' Theorem , the total flux linked with any closed surface is given as the enclosed charge divided by the permittivity of the free space.

$\therefore \: \phi = \dfrac{q}{ \epsilon_{0} }$

Multiplying 4π on both numertoyand denominator:

$= > \: \phi = \dfrac{(4\pi) q}{ 4\pi \epsilon_{0} }$

Putting value of $\dfrac{1}{4\pi\epsilon_{0}}$ as 9 × 10^9

$=> \: \phi = 9\times {10}^{9}\times(4\pi\times 6\times{10}^{-6})$

$= > \: \phi = 216\pi \times {10}^{3}$

$= > \: \phi = 678.5\times {10}^{3}$

$= > \: \phi = 6.78\times {10}^{5} \: \: N {m}^{2} {C}^{-1}$

So, final answer is:

$\boxed{ \sf{\: \phi = 6.78\times {10}^{5} \: \: N {m}^{2} {C}^{-1}}}$