Question

The four small spheres shown in figure
carry charges q; = 4 nC, q2 = 7.8 nC,
q=-2.4 nC and qu= 6 nc
Calculate the flux passing through each of
the three closed surfaces Si, S, and S, in
figure.​

Answer 1

Given that,

Charge$q_{1}=4\ nC$

Charge$q_{2}=7.8\ nC$

Charge$q_{3}=2.4\ nC$

Charge$q_{4}=6\ nC$

According to figure,

The charges in closed surfaces are

$S_{1}=q_{1}$

$S_{2}=q_{3}$

$S_{3}=q_{2}, q_{3}$

We need to calculate the the electric flux through S₁₂₃ closed surface

Using formula of flux

$\phi_{1}=\dfrac{q_{1}}{\epsilon_{0}}$

$\phi_{1}=\dfrac{4\times10^{-9}}{8.85\times10^{-12}}$

$\phi_{1}=451.9\ Nm^2/C$

We need to calculate the the electric flux through S₂ closed surface

Using formula of flux

$\phi_{2}=\dfrac{q_{3}}{\epsilon_{0}}$

$\phi_{2}=\dfrac{-2.4\times10^{-9}}{8.85\times10^{-12}}$

$\phi_{2}=-271.1\ Nm^2/C$

We need to calculate the the electric flux through S₃ closed surface

Using formula of flux

$\phi_{3}=\dfrac{q_{2}+q_{3}}{\epsilon_{0}}$

$\phi_{3}=\dfrac{7.8\times10^{-9}-2.4\times10^{-9}}{8.85\times10^{-12}}$

$\phi_{3}=610.1\ Nm^2/C$

Hence, The flux passing through each of  the three closed surfaces are 451.9 Nm²/C, -271.1  Nm²/C and  610.1 Nm²/C.

Answer 2

Flux through closed surface S

3

,ϕ

S

3

=

ϵ

0

Q

in

=

ϵ

0

q

1

+q

2

=((4.00−7.8)×10

−9

C)/ϵ

0

=−429Nm

2

/C