Lf the electric flux entering and leaving an enclosed surface respectively is -1 and -2- the electric charge inside the surface will be -

The correct option is C (ϕ_2 ϕ_1)ε _0 According Gauss's law, change in flux =Q_inϵ_0 or (ϕ_2 ϕ_1)=Q_inϵ_0 or Q_in=(ϕ_2 ϕ_1)ϵ_0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Gauss's Law

lf the electr...

Question

lf the electric flux entering and leaving an enclosed surface respectively is $ϕ_{1}$ and $ϕ_{2}$ , the electric charge inside the surface will be :

(ϕ_{2} - ϕ_{1}) ε_{0}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{(ϕ_{2} + ϕ_{1})}{ε_{0}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(ϕ_{2} - ϕ_{1})}{ε_{0}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(ϕ_{2} + ϕ_{1}) ε_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(ϕ_{2} - ϕ_{1}) ε_{0}$
According Gauss's law, change in flux $= \frac{Q_{i n}}{ϵ_{0}}$
or $(ϕ_{2} - ϕ_{1}) = \frac{Q_{i n}}{ϵ_{0}}$

or $Q_{i n} = (ϕ_{2} - ϕ_{1}) ϵ_{0}$

Suggest Corrections

Similar questions

Q. If the electric flux entering and leaving an enclosed surface respectively is

ϕ_{1}

and

ϕ_{2}

, the electric charge inside the surface will be

Q. If the electric flux entering and leaving a closed surface are respectively of magnitude

ϕ_{1}

and

ϕ_{2}

then the electric charge inside the surface will be

Q. In the given figure flux through surface

S_{1}

ϕ_{1}

and through

S_{2}

ϕ_{2}

. Which is correct?

Q. The two circular surfaces of a frustum cut out of a right circular cone have radii

2 R

and

3 R

with their centres at

A

and

B

respectively and height,

h = 3 R

. When a point charge

Q

is placed at

A

, the flux of electric field through the circular face of radius

3 R

ϕ_{1}

, and when the charge

Q

is placed at

B

, the flux through the circular face of radius

2 R

ϕ_{2}

.
The ratio

\frac{ϕ_{1}}{ϕ_{2}}

is :

The graph between the stopping potential ( $V_{0}$ ) and $(\frac{1}{λ})$ is shown in the figure. $ϕ_{1}$ , $ϕ_{2}$ and $ϕ_{3}$ are work functions, which of the following is/are correct.

Join BYJU'S Learning Program

lf the electric flux entering and leaving an enclosed surface respectively is ϕ1 and ϕ2, the electric charge inside the surface will be :

The correct option is C (ϕ2−ϕ1)ε0According Gauss's law, change in flux =Qinϵ0or (ϕ2−ϕ1)=Qinϵ0or Qin=(ϕ2−ϕ1)ϵ0

lf the electric flux entering and leaving an enclosed surface respectively is $ϕ_{1}$ and $ϕ_{2}$ , the electric charge inside the surface will be :

The correct option is C $(ϕ_{2} - ϕ_{1}) ε_{0}$
According Gauss's law, change in flux $= \frac{Q_{i n}}{ϵ_{0}}$
or $(ϕ_{2} - ϕ_{1}) = \frac{Q_{i n}}{ϵ_{0}}$

or $Q_{i n} = (ϕ_{2} - ϕ_{1}) ϵ_{0}$