Electric Field and Charge Particle in Electric Field

Q 1 :
An infinitely long positively charged straight thread has a linear charge density $λ$ $C m^{- 1}$ . An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular path from the wire is [2024]

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(2)

Electric field E at a distance r due to infinite long wire is $E = \frac{2 k λ}{r}$

Force of electron, $F = e E \Rightarrow F = e (\frac{2 k λ}{r}) = \frac{2 k λ e}{r}$

This force will provide the required centripetal force.

$F = \frac{m v^{2}}{r} = \frac{2 k λ e}{r} \Rightarrow v = \sqrt{\frac{2 k λ e}{m}}$

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} m (\frac{2 k λ e}{m}) = k λ e = constant$

Q 2 :
$σ$ is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is: [2024]

$\frac{σ}{ε_{0} R}$
$\frac{σ}{ε_{0}}$
$\frac{σ}{2 ε_{0}}$
$\frac{σ}{4 ε_{0}}$

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(2)

By Gauss law $\int \vec{E} \cdot d \vec{A} = \frac{q_{in}}{ε_{0}}$

$E d A = \frac{σ \times d A}{ε_{0}} \Rightarrow E = \frac{σ}{ε_{0}}$

Q 3 :
A particle of charge ‘−q’ and mass ‘m’ moves in a circle of radius ‘r’ around an infinitely long line charge of linear density ‘ $+ λ$ ’. Then time period will be given as

(Consider k as Coulomb’s constant) [2024]

$T^{2} = \frac{4 π^{2} m}{2 k λ q} r^{3}$
$T = 2 π r \sqrt{\frac{m}{2 k λ q}}$
$T = \frac{1}{2 π r} \sqrt{\frac{m}{2 k λ q}}$
$T = \frac{1}{2 π} \sqrt{\frac{2 k λ q}{m}}$

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(2)

$F_{e} = \frac{m v^{2}}{r} \Rightarrow q E = \frac{m v^{2}}{r}$

$q (\frac{2 K λ}{r}) = \frac{m v^{2}}{r} \Rightarrow v = \sqrt{\frac{2 K q λ}{m}}$

$T = \frac{2 π r}{v} = \frac{2 π r}{\sqrt{\frac{2 K q λ}{m}}} = 2 π r \sqrt{\frac{m}{2 K q λ}}$

Q 4 :
Two charges q and 3q are separated by a distance ‘r’ in air. At a distance x from charge q, the resultant electric field is zero. The value of x is [2024]

$\frac{r}{(1 + \sqrt{3})}$
$\frac{(1 + \sqrt{3})}{r}$
$r (1 + \sqrt{3})$
$\frac{r}{3 (1 + \sqrt{3})}$

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(1)

${({\vec{E}}_{net})}_{P} = 0 \Rightarrow \frac{k q}{x^{2}} = \frac{k \cdot 3 q}{{(r - x)}^{2}}$

${(r - x)}^{2} = 3 x^{2}$

$r - x = \sqrt{3} x \Rightarrow x = \frac{r}{\sqrt{3} + 1}$

Q 5 :
If the net electric field at point P along Y axis is zero, then the ratio of $| \frac{q_{2}}{q_{3}} |$ is $\frac{8}{5 \sqrt{x}}$ , where $x$ = ______ . [2024]

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(5)

$\frac{K q_{2}}{{(\sqrt{20})}^{2}} \cos β = \frac{K q_{3}}{{(5)}^{2}} \cos θ$

$\frac{K q_{2}}{20} \frac{4}{\sqrt{20}} = \frac{K q_{3}}{25} \frac{4}{\sqrt{25}}$

$\frac{q_{2}}{q_{3}} = \frac{20}{25} \frac{\sqrt{20}}{\sqrt{25}} = \frac{8}{5 \sqrt{x}}$

$\Rightarrow \sqrt{x} = \frac{8 \times 25 \sqrt{25}}{5 \times 20 \sqrt{20}} \Rightarrow x = 5$

Q 6 :
An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density $+ σ$ . The electron at t = 0 is at a distance of 1 m from S and has a speed of 1 m/s. The maximum value of $σ$ if the electron strikes S at t = 1s is $α [\frac{m ε_{0}}{e}] \frac{C}{m^{2}},$ the value of $α$ is _________ . [2024]

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(8)

Initially the electron must move away from sheet

Electric field due to infinite plane sheet of charge,

$E = \frac{σ}{2 ε_{0}}$

Force experienced by the electron towards the sheet, $F = e E$

Hence, acceleration of the electron, $a = \frac{F}{m}$

$a = \frac{e E}{m} = \frac{e σ}{2 ε_{0} m}$

Using $s = u t + \frac{1}{2} a t^{2}$

$- 1 = 1 \times 1 + \frac{1}{2} (- \frac{e σ}{2 ε_{0} m}) 1^{2}$

$\frac{e σ}{4 ε_{0} m} = 2 \Rightarrow σ = 8 \frac{ε_{0} m}{e}$

Q 7 :
Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^{4}$ N/C normally. A charged particle of mass 2g being suspended through a silk thread of length 20 cm and remains stayed at a distance of 10 cm from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{x}} μ C$ where $x =$ __________. [use g = 10 $m / s^{2}$ ] [2024]

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(3)

$\sin θ = \frac{10}{20} = \frac{1}{2} \Rightarrow θ = 30 °$

Along x-axis

$\sum {(f_{net})}_{y} = 0 (for equilibrium)$

$T \sin θ = q E$ ...(i)

Along y-axis

$\sum {(f_{net})}_{x} = 0 (for equilibrium)$

$T \cos θ = m g$ ...(ii)

From (i) divide by (ii), $\tan θ = \frac{q E}{m g}$

$q = \frac{m g \tan θ}{E}$

$= \frac{2 \times 10^{- 3} \times 10}{2 \times 10^{4}} \times \frac{1}{\sqrt{3}} = \frac{10^{- 6}}{\sqrt{3}} C$

$\Rightarrow q = \frac{1}{\sqrt{3}} μ C \Rightarrow x = 3$

Q 8 :
An electron is made to enters symmetrically between two paralel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the field region with a horizontal component of velocity $10^{6}$ m/s. If the magnitude of the electric field between the plates is 9.1 V/cm, then the vertical component of velocity of electron is (mass of electron $9.1 \times 10^{- 31} kg$ and charge of electron $1.6 \times 10^{- 19}$ ) [2025]

$1 \times 10^{6} m / s$
0
$16 \times 10^{6} m$
$16 \times 10^{4} m / s$

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(3)

Horizontal component of velocity will remain same.

$\Rightarrow t = \frac{l}{V_{x}} = \frac{10 \times 10^{- 2}}{10^{6}} = 10^{- 7}$

$\Rightarrow V_{y} = u_{y} + a_{y} t \Rightarrow V_{y} = 0 + \frac{e E}{m} \times 10^{- 7}$

$\Rightarrow V_{y} = \frac{1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}} \times 9.1 \times 10^{+ 2} \times 10^{- 7}$

$\Rightarrow V_{y} = 16 \times 10^{6} m / s$

Q 9 :
A particle of mass 'm' and charge 'q' is fastened to one end 'A' of a massless string having equilibrium length $l$ , whose other end is fixed at point 'O'. The whole system is placed on a frictiionless horizontal plane and is initialy at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x-axis is [2025]

$\sqrt{\frac{2 q E l}{m}}$
$\sqrt{\frac{q E l}{4 m}}$
$\sqrt{\frac{q E l}{m}}$
$\sqrt{\frac{q E l}{2 m}}$

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(3)

From work energy theorem,

Work done by electric force = change in kinetic energy or

$q E (l - l \cos 60 °) = \frac{1}{2} m v^{2}$

$\Rightarrow \frac{q E l}{2} = \frac{1}{2} m v^{2} or v = \sqrt{\frac{q E l}{m}}$

Q 10 :
A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is: [2025]

(4d/3, 0 , 0)
(d/4, 0 , 0)
(3d/4, 0 , 0)
(d/3, 0 , 0)

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(2)

Figure

Let $E_{P} = 0$

$∴ \frac{k q}{x^{2}} = \frac{k 9 q}{{(d - x)}^{2}} \Rightarrow \frac{d - x}{x} = 3 \Rightarrow x = \frac{d}{4}$

co-cordinate of P is $(\frac{d}{4}, 0, 0)$

Topic Question Set

Q 2 :
$σ$ is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is: [2024]

Q 3 :
A particle of charge ‘−q’ and mass ‘m’ moves in a circle of radius ‘r’ around an infinitely long line charge of linear density ‘ $+ λ$ ’. Then time period will be given as

(Consider k as Coulomb’s constant) [2024]

Q 4 :
Two charges q and 3q are separated by a distance ‘r’ in air. At a distance x from charge q, the resultant electric field is zero. The value of x is [2024]

Q 5 :
If the net electric field at point P along Y axis is zero, then the ratio of $| \frac{q_{2}}{q_{3}} |$ is $\frac{8}{5 \sqrt{x}}$ , where $x$ = ______ . [2024]

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Q 10 :
A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is: [2025]

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Topic Question Set

Q 2 : σ is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is: [2024]

Q 3 : A particle of charge ‘−q’ and mass ‘m’ moves in a circle of radius ‘r’ around an infinitely long line charge of linear density ‘+λ’. Then time period will be given as (Consider k as Coulomb’s constant) [2024]

Q 4 : Two charges q and 3q are separated by a distance ‘r’ in air. At a distance x from charge q, the resultant electric field is zero. The value of x is [2024]

Q 5 : If the net electric field at point P along Y axis is zero, then the ratio of |q2q3| is 85x, where x = ______ . [2024]

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Q 10 : A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is: [2025]

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Q 2 :
$σ$ is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is: [2024]

Q 3 :
A particle of charge ‘−q’ and mass ‘m’ moves in a circle of radius ‘r’ around an infinitely long line charge of linear density ‘ $+ λ$ ’. Then time period will be given as

(Consider k as Coulomb’s constant) [2024]

Q 4 :
Two charges q and 3q are separated by a distance ‘r’ in air. At a distance x from charge q, the resultant electric field is zero. The value of x is [2024]

Q 5 :
If the net electric field at point P along Y axis is zero, then the ratio of $| \frac{q_{2}}{q_{3}} |$ is $\frac{8}{5 \sqrt{x}}$ , where $x$ = ______ . [2024]

Q 10 :
A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is: [2025]