Question

$\large \underline\pink{\bf\pmb{Question}}$
A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through two opposite faces of the cube? (All India)
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Answer 1

Answer:

Side length of the cube = l

The magnitude of the charge placed at the center of cube = q

To Find :

Magnitude of electric flux passing through two opposite faces of the cube = ?

Solution :

According to Gauss's law , we know that formula for electric flux is given as :

\phi = \frac{q}{\epsilon_o}ϕ=

ϵ

o

q

Here , q is total charge enclosed within the closed surface

And \epsilon_0ϵ

0

is permitivity of free space

Now for a charge q inside a cube the flux through each face is :

= \frac{q}{6\epsilon_o}

6ϵ

o

q

So flux passing through two opposite faces will be :

= 2 \times \frac{q}{6 \epsilon_o}=2×

6ϵ

o

q

= \frac{q}{3 \epsilon_o}=

3ϵ

o

q

So the electric flux passing through two opposite faces of the cube is = \frac{q}{3 \epsilon_o}=

3ϵ

o

Answer 2

Answer:

hi sis here is your answer

By symmetry, the flux through each of the six faces of the cube will be same when charge q is placed at its centre. Thus, electric flux passing through two opposite faces of the cube = 2.1/6.q/ε0 .

hope it helps you

thanks dedo please