Q: A cylinder of radius \$R\$ and length \$L\$ is placed in a uniform electric field \$E\$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by :

Flux through surface \$A\$,\${ \phi }_{ A }=E\times \pi { R }^{ 2 }\$\${ \phi }_{ B }=-E\times \pi { R }^{ 2 }\$Flux through curved surface, \$C=\displaystyle\int { E.ds } \$ \$=\displaystyle\int { Eds } \cos { { 90 }^{ o } } =0\$\$\therefore \$ Total flux through cylinder \$={ \phi }_{ A }+{ \phi }_{ B }+{ \phi }_{ C }=0\$

Q: If the electric flux entering and leaving an enclosed surface respectively are \$\phi_1\$ and \$\phi_2\$, the electric charge inside the surface will be :

Hint: Here, we have to apply gauss's law to find electric charge. Solution:We know that, electric flux \$\phi_{1}\$ (or electric field lines) entering in a closed surface is -ve and electric flux \$\phi_{2}\$ (or electric field lines) leaving a closed surface is +ve. Hence, net electric flux through the closed surface, \$\phi=\phi_{2}-\phi_{1}\$Now, according to Gauss' theorem, the net electric flux \$\phi\$ passing through a closed surface is equal to the \$1/\varepsilon_{0}\$ times of the total charge \$q\$, inside the surface. Step1: Apply gauss's lawGiven, Net electric flux, \$\phi = ({\phi _2} - {\phi _1})\$ \$ \phi = \dfrac{q_{in}}{{{\varepsilon _0}}}\$ \$\Rightarrow q_{in} = {\varepsilon _0}\phi \$ \$\therefore q_{in} = ({\phi _2} - {\phi _1}){\varepsilon _0}\$ Hence option ‘D’ is correct.

Q: Figure shows three point charges \$+2\ q,-q\ and +3\ q\$. Two charges \$+2\ q\ and q\$ are enclosed within a surface \$S\$.What is the electric flux due to this configuration through the surface \$S\$?

In given figure,\$\begin{array}{l} qin=+2q-q \\ \, \, \, \, \, \, \, \, =q \\ \therefore \phi =\frac { q }{ { { E_{ 0 } } } } \end{array}\$

Q: A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will :

\$\large \textbf{Method 1: Using Number of Field Lines}\$\$\bullet\$ We know that, Electric flux is proportional to number of field lines.\$\bullet\$ Form figure, we see that the number of field lines crossing both the spherical surfaces is same.\$\bullet\$ Therefore, the flux passing both the surfaces will be same.Hence correct Option is \$D\$\$\large \textbf{Method 2: Using Gauss Law}\$\$\bullet\$ Gauss Law states that, Electric flux \$\Delta \phi = \cfrac{q_{in}}{\epsilon_0}\$\$\bullet\$ Assuming both the spherical surfaces \$S_1\$ and \$S_2\$ as gaussian surfaces. From figure, we find that the charge enclosed(\$q_{in}=Q\$), same for both the surfaces irrespective of radius.\$\bullet\$ Hence, the Flux passing through both surfaces is same.Hence correct Option is \$D\$\$\textbf{Note: }\$ We can observe that whatever be the size, shape and position of second surface, If it is enclosing the same charge, then the Number of field lines crossing it or charge enclosed by it remains the same as first surface. Hence answer will be same, even if the second surface is cube or cylinder of any size.

Q: A square surface of side L meter in the plane of the paper is placed in a uniform electric field E(volt/m) acting along the same plane at an angle with the horizontal side of the square as shown in figure. The electric flux linked to the surface,in units of volt\$-m\$, is

The electric flux linked to the surface is given by:\$\phi_E=\overrightarrow E.d\overrightarrow A\$The angle between the area vector and the field lines is \$90^o\$.Thus,\$\phi_E=EAcos\theta = 0\$Also,As, there is no field line crossing the surface, the electric flux linked to the surface will be zero.

Q: The electric field in a region of space is given by E = 5i +2j N/C. The electric flux due to this field through an area \$2{m^2}\$ lying in the YZ plane, in S.I unit, is

Only the \$'i'\$ component is E will cause any flux through the y plane. This 'j' component is parallel to the y,z plane so will not push anything through it.Hence the flux is\$\Rightarrow \phi=5(N/C)(2m^2)=10\$

Q: The magnetic flux through a coil perpendicular to the plane is varying according to the relation \$\varphi =\left( 5{ t }^{ 3 }+4{ t }^{ 2 }+2t-5 \right) \$ wb. Calculate the induced current through the coil at \$t=2\ sec\$, if the resistance of the coil is \$5\ \Omega \$.

\$\phi =(5t^3+4t^2+2t-5)\$ Webertime \$t=2\ sec\$Resistance \$=5\Omega\$induced emf \$E=\dfrac{d\phi}{dt}\$\$\dfrac{d\phi}{dt}=15t^2+8t+2\$at \$t=2\ sec, \dfrac{d\phi}{dt}=15\times 4+8\times 2+2\$ \$=78\ V\$Also \$E=IR\$\$E=78\ V, R=5\Omega\$\$I=\dfrac{k}{R}=\dfrac{78}{5}=15.6\ A\$

Question 1

A cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by :

Accepted Answer

Flux through surface $A$,${ \phi  }_{ A }=E\times \pi { R }^{ 2 }$${ \phi  }_{ B }=-E\times \pi { R }^{ 2 }$Flux through curved surface, $C=\displaystyle\int { E.ds } $                                                   $=\displaystyle\int { Eds } \cos { { 90 }^{ o } } =0$$\therefore $ Total flux through cylinder $={ \phi  }_{ A }+{ \phi  }_{ B }+{ \phi  }_{ C }=0$

Question 2

If the electric flux entering and leaving an enclosed surface respectively are $\phi_1$ and $\phi_2$, the electric charge inside the surface will be :

Accepted Answer

Hint: Here, we have to apply gauss's law to find electric&#10;charge.&#10;&#10;Solution:We know that, electric flux $\phi_{1}$ (or electric field lines) entering in a closed surface is -ve and electric flux $\phi_{2}$ (or electric field lines) leaving a closed surface is +ve. Hence, net electric flux through the closed surface,                    $\phi=\phi_{2}-\phi_{1}$Now, according to Gauss' theorem, the net electric flux $\phi$ passing through a closed surface is equal to the $1/\varepsilon_{0}$ times of the total charge $q$, inside the surface.&#10;&#10;Step1: Apply gauss's lawGiven, Net electric flux, $\phi  = ({\phi _2} - {\phi _1})$&#10;&#10;$ \phi  = \dfrac{q_{in}}{{{\varepsilon _0}}}$&#10;&#10;$\Rightarrow q_{in} =&#10;{\varepsilon _0}\phi $&#10;&#10;$\therefore q_{in} = ({\phi _2}&#10;- {\phi _1}){\varepsilon _0}$&#10;&#10;Hence option &#8216;D&#8217; is correct.&#10;&#10; &#10;&#10;

Question 3

Figure shows three point charges $+2\ q,-q\ and +3\ q$. Two charges $+2\ q\ and q$ are enclosed within a surface $S$.What is the electric flux due to this configuration through the surface $S$?

Accepted Answer

In given figure,$\begin{array}{l} qin=+2q-q \ \, \, \, \, \, \, \, \, =q \ \therefore \phi =\frac { q }{ { { E_{ 0 } } } }  \end{array}$

Question 4

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will :

Accepted Answer

$\large \textbf{Method 1: Using Number of Field Lines}$$\bullet$  We know that, Electric flux is proportional to number of field lines.$\bullet$  Form figure, we see that the number of field lines crossing both the spherical surfaces is same.$\bullet$  Therefore, the flux passing both the surfaces will be same.Hence correct Option is $D$$\large \textbf{Method 2: Using Gauss Law}$$\bullet$  Gauss Law states that, Electric flux $\Delta \phi = \cfrac{q_{in}}{\epsilon_0}$$\bullet$  Assuming both the spherical surfaces $S_1$ and $S_2$ as gaussian surfaces. From figure, we find that the charge enclosed($q_{in}=Q$), same for both the surfaces irrespective of radius.$\bullet$  Hence, the Flux passing through both surfaces is same.Hence correct Option is $D$$\textbf{Note: }$ We can observe that whatever be the size, shape and position of second surface, If it is enclosing the same charge, then the Number of field lines crossing it or charge enclosed by it remains the same as first surface. Hence answer will be same, even if the second surface is cube or cylinder of any size.

Question 5

A square surface of side L meter in the plane of the paper is placed in a uniform electric field E(volt/m) acting along the same plane at an angle with the horizontal side of the square as shown in figure. The electric flux linked to the surface,in units of volt$-m$, is

Accepted Answer

The electric flux linked to the surface is given by:$\phi_E=\overrightarrow E.d\overrightarrow A$The angle between the area vector and the field lines is $90^o$.Thus,$\phi_E=EAcos\theta = 0$Also,As, there is no field line crossing the surface, the electric flux linked to the surface  will be zero.

Question 6

A charge $q$ is placed at the corner of a cube of side $a$. The electric flux passing through the cube is :

Accepted Answer

According to Gauss's law, the electric flux through a closed surface is equal to $\cfrac { 1 }{ { \varepsilon  }_{ 0 } } $ times the net charge enclosed by the surface.Since, $q$ is the charge enclosed by the surface, then the electric flux $\phi =\cfrac { q }{ { \varepsilon  }_{ 0 } } $.If charge $q$ is placed at a corner of cube, it will be divided ito 8 such cubes. Therefore, electric flux through the cube is$\phi '=\cfrac { 1 }{ 8 } \left( \cfrac { q }{ { \varepsilon  }_{ 0 } }  \right) $

Question 7

The electric field in a region of space is given by E = 5i +2j N/C. The electric flux due to this field through an area $2{m^2}$ lying in the YZ plane, in S.I unit, is

Accepted Answer

Only the $'i'$ component is E will cause any flux through the y plane. This 'j' component is parallel to the y,z plane so will not push anything through it.Hence the flux is$\Rightarrow \phi=5(N/C)(2m^2)=10$

Question 8

Define electric flux. Write its SI unit.

Accepted Answer

Electric flux is defined as the measure of count of number of electric field lines crossing an area. Electric flux   $\phi = E A\cos\theta$SI unit of electric flux is  $Nm^2/C$

Question 9

Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be ?

Accepted Answer

The total flux linked with the closed surface is given by Gauss's law,$\phi=\int \vec E.\vec {ds}=\dfrac{Q_{encl}}{\varepsilon_0}$So, the electric field will be due to all the charge enclosed by the surface.The correct option is C.

Question 10

The magnetic flux through a coil perpendicular to the plane is varying according to the relation $\varphi =\left( 5{ t }^{ 3 }+4{ t }^{ 2 }+2t-5 \right) $ wb. Calculate the induced current through the coil at $t=2\ sec$, if the resistance of the coil is $5\ \Omega $.

Accepted Answer

$\phi =(5t^3+4t^2+2t-5)$ Webertime $t=2\ sec$Resistance $=5\Omega$induced emf $E=\dfrac{d\phi}{dt}$$\dfrac{d\phi}{dt}=15t^2+8t+2$at $t=2\ sec, \dfrac{d\phi}{dt}=15\times 4+8\times 2+2$                              $=78\ V$Also $E=IR$$E=78\ V, R=5\Omega$$I=\dfrac{k}{R}=\dfrac{78}{5}=15.6\ A$

Solid Angle

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REVISE WITH CONCEPTS

Area Vector, Solid Angle and Electric Flux

QUICK SUMMARY WITH STORIES

Introduction to Electric Flux

A cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by :

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

Figure shows three point charges $+ 2 q, - q a n d + 3 q$ . Two charges $+ 2 q a n d q$ are enclosed within a surface $S$ .
What is the electric flux due to this configuration through the surface $S$ ?

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will :

A square surface of side L meter in the plane of the paper is placed in a uniform electric field E(volt/m) acting along the same plane at an angle with the horizontal side of the square as shown in figure. The electric flux linked to the surface,in units of volt $- m$ , is

A charge $q$ is placed at the corner of a cube of side $a$ . The electric flux passing through the cube is :

The electric field in a region of space is given by E = 5i +2j N/C. The electric flux due to this field through an area $2 m^{2}$ lying in the YZ plane, in S.I unit, is

Define electric flux. Write its SI unit.

Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be ?

The magnetic flux through a coil perpendicular to the plane is varying according to the relation $φ = (5 t^{3} + 4 t^{2} + 2 t - 5)$ wb. Calculate the induced current through the coil at $t = 2 s e c$ , if the resistance of the coil is $5 Ω$ .

Solid Angle

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REVISE WITH CONCEPTS

Area Vector, Solid Angle and Electric Flux

QUICK SUMMARY WITH STORIES

Introduction to Electric Flux

A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by :

If the electric flux entering and leaving an enclosed surface respectively are ϕ1​ and ϕ2​, the electric charge inside the surface will be :

Figure shows three point charges +2 q,−q and+3 q. Two charges +2 q andq are enclosed within a surface S. What is the electric flux due to this configuration through the surface S?

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will :

A square surface of side L meter in the plane of the paper is placed in a uniform electric field E(volt/m) acting along the same plane at an angle with the horizontal side of the square as shown in figure. The electric flux linked to the surface,in units of volt−m, is

A charge q is placed at the corner of a cube of side a. The electric flux passing through the cube is :

The electric field in a region of space is given by E = 5i +2j N/C. The electric flux due to this field through an area 2m2 lying in the YZ plane, in S.I unit, is

Define electric flux. Write its SI unit.

Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be ?

The magnetic flux through a coil perpendicular to the plane is varying according to the relation φ=(5t3+4t2+2t−5) wb. Calculate the induced current through the coil at t=2 sec, if the resistance of the coil is 5 Ω.

A cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by :

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

Figure shows three point charges $+ 2 q, - q a n d + 3 q$ . Two charges $+ 2 q a n d q$ are enclosed within a surface $S$ .
What is the electric flux due to this configuration through the surface $S$ ?

A square surface of side L meter in the plane of the paper is placed in a uniform electric field E(volt/m) acting along the same plane at an angle with the horizontal side of the square as shown in figure. The electric flux linked to the surface,in units of volt $- m$ , is

A charge $q$ is placed at the corner of a cube of side $a$ . The electric flux passing through the cube is :

The electric field in a region of space is given by E = 5i +2j N/C. The electric flux due to this field through an area $2 m^{2}$ lying in the YZ plane, in S.I unit, is

The magnetic flux through a coil perpendicular to the plane is varying according to the relation $φ = (5 t^{3} + 4 t^{2} + 2 t - 5)$ wb. Calculate the induced current through the coil at $t = 2 s e c$ , if the resistance of the coil is $5 Ω$ .