# Q 1.17 (a) Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^{3}\frac{Nm^{2}}{C}$. (a) What is the net charge inside the box?

Using Gauss’s law, we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$  times the total charge enclosed by S.

i.e.   $\phi = q/\epsilon _{0}$

where, q = net charge enclosed  and  $\epsilon _{0}$ = permittivity of free space (constant)

Given, $\phi = 8.0 \times10^3\ Nm^2/C$

$\therefore q = \phi \times\epsilon _{0} = (8.0\times10^3 \times8.85\times10^{-12})\ C$

$\implies q = 70 \times10^{-9} C = \boldsymbol{0.07 \mu C}$

This is the net charge inside the box.

## Related Chapters

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-