An electron moves with a velocity \$\vec{v}\$ in an electric field \$\vec{E}\$. If the angle between \$\vec{v}\$ and \$\vec{E}\$ is neither 0 nor \$\pi \$, then the path followed by the electron is a(an):

The perpendicular component of velocity is given by:\$v_y=v \sin\theta\$Now in the X direction, the electron experiences a force of \$E(-e)\$So, \$F= -eE\$or, \$a =\dfrac {-Ee}{m}\$\$\therefore\$ velocity in X direction: \$v_x= u + at \$ \$= v cos\theta-\dfrac {Ee}{m}\times t\$We know, \$ y=\int v_ydt\$ and \$x=\int (v_x)dt\$\$\Rightarrow\$ \$y=v \sin\theta \cdot t, x=v cos\theta \cdot t-\dfrac {Ee}{m}\dfrac {t^2}{2}\$Substituting \$t=\dfrac {y}{v sin\theta}\$, we get \$x=v \cos\theta \dfrac { y}{v sin\theta}-\dfrac {Ee}{m_2}\cdot \dfrac {y^2}{v^2 sin^2\theta}\$\$\Rightarrow\$ \$x=y \cot\theta -\dfrac {Ee}{2m v^2sin^2\theta}y^2\$, where \$\theta\$ is the initial angle of \$v\$ with \$E\$.\$\therefore\$ path followed is a parabola

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>>Class 12
>>Physics
>>Electric Charges and Fields
>> $Medium Questions$

Electric Charges And Fields

Q: A large flat metal surface has uniform charge density \$+\sigma \$. An electron of mass \$m\$ and charge \$e\$ leaves the surface at point A with speed \$v\$, and return to it at point B. The maximum value of AB is:

Field near metal surface \$ E = \displaystyle\frac{\sigma}{\epsilon_o} \$Force on electron \$ = eE = \displaystyle\frac{e \sigma}{\epsilon_o} \$Acceleration of electron \$ = a = \displaystyle\frac{e \sigma}{m \epsilon_o} \$Max Range \$ = \displaystyle\frac{v^2}{a} = \displaystyle\frac{v^2 m \epsilon_o}{e \sigma} \$

Q: A cube is arranged such that its length, breadth, height are along X,Y and Z directions. One of its corners is situated at the origin. Length of each side of the cube is \$25\ cm\$ . The components of electric field are \$E_{x}=400\sqrt{2}\ N/C,\ E_{y}=0\$ and \$E_{Z}=0\$ respectively. The flux coming out of the cube at one end, whose plane is perpendicular to X axis, will be:

Flux coming out on one side is \$\vec E.d \vec a\$So flux coming out of the cube through the surface normal to X-axis is given by:\$ \phi =E_x \times (25\times 25\times 10^{-4})\$ \$ =\dfrac{400 \sqrt{2}\times 25\times 25}{10000}\$\$ =25\sqrt{2}\$ \$Nm^2/C\$

Q: Two electric dipoles each of dipole moment \$p=6.2\times 10^{^{-30}}C-m\$ are placed with their axis along the same line and their centres at a distance d \$=10^{-8}cm\$. The force of attraction between dipoles is:

Electric field due to a dipole on its axis is given by \$ 2\overrightarrow{p}/(4\pi\epsilon_0 r^3) \$Therefore using necessary approximation and \$F = qE \$ we get,\$ F = 2qp/(4\pi\epsilon_0) (1/(d-l)^3 - 1/(d+l)^3 ) = 2qp/(4\pi\epsilon_0) 6d^2l/d^6 = 6p^2 / 4\pi\epsilon_0 d^4 \$putting in the values,\$ F = 6\times 9\times 10^9\times 6.2^2 \times 10^{-60} \times 10^{40} = 2.1 \times 10^{-8} N\$

Q: The electric field in a region of space is given by \$\vec{E}=(\hat{5i}+\hat{2j} )Nc^{-1}\$ . The electric flux due to this field through an area \$2m^{2}\$ lying in the Y-Z plane in S.I. units is

As \$2m^{2}\$ area is in y-z plane its area vector which is perpendicular to y-z plane is in x axis.\$\therefore\$ area vector \$=2(\hat{i})\$given \$E=5i+2j\$We know flux \$\phi=E.A\$\$=(2i).(5i+2j)\$\$\phi=10\$

Q: Drawings I and II show two samples of electric field lines. Then :

The magnitude of electric field at a point can be determined by knowing how closely are the electric field lines packed at that particular point.In case 1: the field lines are equally spaced everywhere.In case 2: the field lines are spaced more closely in the right than in the left.

Q: A particle that carries a charge \$-q\$ is placed at rest in uniform electric field \$10\ N/C\$. It experiences a force and moves in a certain time t, it is observed to acquire a velocity \$10\vec{i}-10\vec{j}\$ m/s. The given electric field intersects a surface of area \$A\$ \$m^{2}\$in the X -Z plane. Electric flux through surface is:

\$v=10\hat{i}-10\hat{j}\$As velocity is in x and y components Electric field also has two components Let \$E=Ex\hat{i}+Ey\hat{j}\$Now force on \$(-q)=E(-q)\$Force \$F=-Exq\hat{i}+-Eyq\hat{j}\$Acceleration \$a =(-\dfrac{Exq}{m})\hat{i} +(-\dfrac{EyV}{m})\hat{i}\$Now \$v_x=10\$ and \$v_y=-10\$we know \$v =u+at \ \ (given \ u=0 )\$\$\therefore V=at\$\$\Rightarrow v_x=(-\dfrac{Exq}{m})t --(1) \ \ and\ \ v_y=(\dfrac{-Eyq}{m})t --(2)\$\$Given |E|=10\ N/C\$Now from 1 and 2 we get, \$-1=\dfrac{-Exq}{m}\dfrac{m}{Eyq}\$\$\Rightarrow Ex=Ey \$\$10=\sqrt{Ex^{2}+Ey^{2}}\$\$\Rightarrow Ex=5\sqrt{2}=Ey\$Now than \$\sigma = \int E.dA\$ As the area is in X-Z plane Area =\$A\$\$\hat{i}\$\$\therefore Ea=Ey.A\$\$\phi=5\sqrt{2}A\$

Q: The dipolemoment of the given system is:

See the diagram.Dipole moment of a system of charges is given by,\$ \overrightarrow p = \Sigma q_i \overrightarrow{r_i} \$\$\vec P = -ql\hat{i} + ql\hat{i} + (-2q)(\sqrt{3}/2\times l )\hat{j} = -\sqrt{3} ql \hat{j} \$

Q: The electric dipole moment is:

We know electric dipole moment\$P= q .d\$where q is the magnitude of each charge d is the distance between themOn putting the values we get\$P=3.2\times 10^{-19}\times2.44\$\$P=7.68\times 10^{-29}\ C-m\$

Q: An electric dipole is placed in a non uniform electric field increasing along the \$+ve\$ direction of X - axis. The dipole moves along \$\underline{\hspace{0.5in}}\$ and rotates \$\underline{\hspace{0.5in}}\$

As electric field is more towards positive side of x axis force acting on +q greater than force acting on -q.total force \$F_{net}=q(E_1(x)-E_2(x)) \ where\ E_1> E_2 \ are\ different\$\$\therefore \$ the dipole will move along the +ve x axis.Torque is given as \$P\times E\$As shown in the diagram the dipole will rotate in clockwise direction.

Physics
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JEE Mains JEE Advanced AIIMS BITSAT Easy Qs Med Qs Hard Qs

A large flat metal surface has uniform charge density $+ σ$ . An electron of mass $m$ and charge $e$ leaves the surface at point A with speed $v$ , and return to it at point B. The maximum value of AB is:

\frac{v m ϵ _{0}}{σ e}

\frac{v ^{2} m ϵ _{0}}{σ e}

\frac{v ^{2} e}{ϵ _{0} σ m}

\frac{v ^{2} σ e}{ϵ _{0} m}

Medium

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A cube is arranged such that its length, breadth, height are along X,Y and Z directions. One of its corners is situated at the origin. Length of each side of the cube is $25 c m$ . The components of electric field are $E_{x} = 4002 N / C, E_{y} = 0$ and $E_{Z} = 0$ respectively. The flux coming out of the cube at one end, whose plane is perpendicular to X axis, will be:

252 N m^{2} / C

52 N m^{2} / C

2502 N m^{2} / C

25 N m^{2} / C

Medium

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Two electric dipoles each of dipole moment $p = 6.2 \times 1 0^{^{- 30}} C - m$ are placed with their axis along the same line and their centres at a distance d $= 1 0^{- 8} c m$ . The force of attraction between dipoles is:

2.1 \times 1 0^{- 16} N

2.1 \times 1 0^{- 12} N

2.1 \times 1 0^{- 10} N

2.1 \times 1 0^{- 8} N

Medium

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An electron moves with a velocity $v$ in an electric field $E$ . If the angle between $v$ and $E$ is neither 0 nor $π$ , then the path followed by the electron is a(an):

straight line

circle

ellipse

parabola

Medium

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The electric field in a region of space is given by $E = (\hat{5 i} + \hat{2 j}) N c^{- 1}$ . The electric flux due to this field through an area $2 m^{2}$ lying in the Y-Z plane in S.I. units is

10

20

102

229

Medium

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Drawings I and II show two samples of electric field lines. Then :

the electric fields in both I and II are produced by negative charge located somewhere on the left and positive charges located somewhere on the right

in both I and II the electric field is the same everywhere

in both cases the field becomes stronger on moving from left to right

the electric field in I is the same everywhere, but in II the electric field becomes stronger on moving from left to right

Medium

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A particle that carries a charge $- q$ is placed at rest in uniform electric field $10 N / C$ . It experiences a force and moves in a certain time t, it is observed to acquire a velocity $10 i - 10 j$ m/s. The given electric field intersects a surface of area $A$ $m^{2}$ in the X -Z plane. Electric flux through surface is:

52 A N m^{2} / C

5 A N m^{2} / C

2 A N m^{2} / C

25 A N m^{2} / C

Medium

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The dipolemoment of the given system is:

3

q l

along perpendicular bisector of

q - q

line

2 q l

along perpendicular bisector of

q - q

line

q l

2

along perpendicular bisector of

q - q

line

0

Medium

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The electric dipole moment is:

15.36 \times 1 0^{- 29} C - m

15.36 \times 1 0^{- 19} C - m

7.68 \times 1 0^{- 29} C - m

7.68 \times 1 0^{- 19} C - m

Medium

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An electric dipole is placed in a non uniform electric field increasing along the $+ v e$ direction of X - axis. The dipole moves along and rotates

+ v e

direction of X - axis, clockwise

- v e

direction of X - axis, clockwise

+ v e

direction of X - axis, anti clockwise

- v e

direction of X - axis, anti clockwise

Medium

View solution>

Electric Charges And Fields

Physics 4588 Views

A large flat metal surface has uniform charge density +σ. An electron of mass m and charge e leaves the surface at point A with speed v, and return to it at point B. The maximum value of AB is:

Two electric dipoles each of dipole moment p=6.2×10−30C−m are placed with their axis along the same line and their centres at a distance d =10−8cm. The force of attraction between dipoles is:

An electron moves with a velocity v in an electric field E. If the angle between v and E is neither 0 nor π, then the path followed by the electron is a(an):

The electric field in a region of space is given by E=(5i^+2j^​)Nc−1 . The electric flux due to this field through an area 2m2 lying in the Y-Z plane in S.I. units is

Drawings I and II show two samples of electric field lines. Then :

The dipolemoment of the given system is:

The electric dipole moment is:

An electric dipole is placed in a non uniform electric field increasing along the +ve direction of X - axis. The dipole moves along ​ and rotates ​

Physics
4588 Views

A large flat metal surface has uniform charge density $+ σ$ . An electron of mass $m$ and charge $e$ leaves the surface at point A with speed $v$ , and return to it at point B. The maximum value of AB is:

Two electric dipoles each of dipole moment $p = 6.2 \times 1 0^{^{- 30}} C - m$ are placed with their axis along the same line and their centres at a distance d $= 1 0^{- 8} c m$ . The force of attraction between dipoles is:

An electron moves with a velocity $v$ in an electric field $E$ . If the angle between $v$ and $E$ is neither 0 nor $π$ , then the path followed by the electron is a(an):

The electric field in a region of space is given by $E = (\hat{5 i} + \hat{2 j}) N c^{- 1}$ . The electric flux due to this field through an area $2 m^{2}$ lying in the Y-Z plane in S.I. units is

An electric dipole is placed in a non uniform electric field increasing along the $+ v e$ direction of X - axis. The dipole moves along and rotates