datavorous/entrance-exam-dataset huggingface dataset:
The electric field in a space Eâ=4i^-5k^ V m-1. What is the potential difference VB-VA between points A and B? The coordinates of A and B are 4, 0, 3 m and 10, 3, 0 m, respectively.

Options:
A -39 V
B 11 V
C 25 V
D -25 V

The correct answer is:
-39 V

Solution:
Since, we know,
V2-V1=-∫r1r2E→⋅dr
Given, E→=4i^-5k^ V m-1,
A→=4i^+3k^ and B→=10i^+3j^
∴ Potential difference =VB-VA
Putting given values, we get
= -∫rArB4i^-5k^⋅dxi^+drj^+dzk^
= -∫4, 0, 310, 3, 04dx-5dz=-4x410-5z30
=-39 V

Topic: Physics, Electrostatics, JEE Main
The electric field in an electromagnetic wave is given as
E→=20sinωt-xcj→N C-1, where ω and c are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×10-4 m3 will be
(Given ε0=8.85×10-12C2 N-1 m-2)

Options:
A 88.5×10-13 J
B 17.7×10-13 J
C 28.5×10-13 J
D 8.85×10-13 J

The correct answer is:
8.85×10-13 J

Solution:
Given that, E→=20sinωt-xcj→N C-1

The average energy density of an electromagnetic wave is 12ε0E20.

The total energy stored is

Etotal =12ε0E02× volume

⇒Etotal=12×8.85×10-12×(20)2×5×10-4=8.85×10-13 J

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The electric field in an electromagnetic wave is given by $\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$. The magnetic field induction of this wave is (in SI unit) :

Options:
A $\overrightarrow{\mathrm{B}}=\hat{k} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$
B $\vec{B}=\hat{j} 40 \cos \omega(t-z / c)$
C $\overrightarrow{\mathrm{B}}=\hat{i} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$
D $\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$

The correct answer is:
$\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$

Solution:
$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$
$\overrightarrow{\mathrm{E}}$ is along $+\mathrm{x}$ direction
$\overrightarrow{\mathrm{v}}$ is along $+\mathrm{z}$ direction
So direction of $\vec{B}$ will be along +y and magnitude of $B$ will be $\frac{E}{c}$
So answer is $\frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right) \hat{j}$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (04 Apr Shift 1)
The electric field intensity at a large distance \( x \) from an electric dipole is proportional to

Options:
A \( \frac{1}{x} \)
B \( \frac{1}{x^{2}} \)
C \( \frac{1}{x^{3}} \)
D \( \frac{1}{x^{4}} \)

The correct answer is:
\( \frac{1}{x^{3}} \)

Solution:
In case of electric dipole, Electric field intensity in axial position is expressed as E=14πϵ0·2pr3

∴ For distance x, E=14πεo·2px3⇒E∝1x3

At a point far from a dipole in any direction, the magnitude of the electric field is proportional to the magnitude of the dipole moment and decreases with the cube of the distance.

Topic: Physics, Electrostatics, JEE Main
The electric field intensity at all points in space is given by Eâ=3i^âj^ Vm-1. The nature of equipotential lines in x-y plane, is shown by

Options:
A
B
C
D

The correct answer is:

Solution:
The direction of uniform electric field in x-y plane is as shown

The equipotential lines will be perpendicular to electric field. Also electric field points from high potential region towards low potential region. Therefore nature of equipotential lines in x-y plane is as shown in option 3

Topic: Physics, Electrostatics, JEE Main
The electric field intensity is produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60 W at the same distance is x5E. Where the value of x=_________ .

Options:

The correct answer is:
3

Solution:
cε0E2=1004π×32

⇒cε0x5E2=604π×32

⇒x5=35

⇒x=3

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The electric field of a plane electromagnetic wave is given by
E→=E0i^cos⁡kzcos⁡(ωt)
The corresponding magnetic field B→ is then given by:

Options:
A B→=E0Cj^cos⁡kzsin(ωt)
B B→=E0Ck^sin⁡kzcos(ωt)
C B→=E0Cj^sin⁡kzsin(ωt)
D B→=E0Cj^sin⁡kzcos(ωt)

The correct answer is:
B→=E0Cj^sin⁡kzsin(ωt)

Solution:
E→=E0 i^ cos kz cosωt
E→=E02 i^ cos kz-ωt+E02i^ coskz+ωt
B→=E02C j^ cos kz-ωt+E02Cj^ coskz+ωt
=E02Cj^ 2 sin kzsinωt
=E0Cj^  sin kzsinωt

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The electric field of a plane electromagnetic wave is given by E→=E0i^+j^2cos⁡kz+ωt . At t=0 , a positively charged particle is at the point x,y,z=0,0,πk . If its instantaneous velocity at t=0 is v0k^ , the force acting on it due to the wave is:

Options:
A parallel to i^+j^2
B zero
C antiparallel to i^+j^2
D parallel to k^

The correct answer is:
antiparallel to i^+j^2

Solution:
Force due to electric field is in direction –i^+j^2E0
Force due to magnetic filed is in direction qv→×B→ and v→||k^
∴ it is parallel to E→
∴ net force is antiparallel to i^+j^2 .

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The electric field of a plane electromagnetic wave is given by E→=E0x^+y^sinkz-ωt. Its magnetic field will be given by

Options:
A E0c-x^+y^sinkz-ωt
B E0cx^+y^sinkz-ωt
C E0cx^-y^sinkz-ωt
D E0cx^-y^coskz-ωt

The correct answer is:
E0c-x^+y^sinkz-ωt

Solution:
As in electromagnetic wave electric field is always perpendicular to the magnetic field and both the fields oscillates perpendicularly to the direction pf propagation so E→×B→‖C→

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is E→=E0jcos(ωt-kx). The magnetic field B→, at the moment t=0 is:

Options:
A B→=E0μ0ϵ0cos(kx)k^
B B→=E0μ0ϵ0cos(kx)j^
C B→=E0μ0ε0cos(kx)k^
D B→=E0μ0ϵ0cos(kx)j^

The correct answer is:
B→=E0μ0ε0cos(kx)k^

Solution:
B0=E0C=E01/μ0ε0=E0μ0ε0

As the light is propagating in x direction

and E^×B^ || C^

∴ B→ should be in K^ direction

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The electric field of a plane polarized electromagnetic wave in free space at time t=0 is given by the expression E→x, y=10j^cos6x+8z. The magnetic field B→x, z, t is given by (c is the velocity of light.)

Options:
A 1c6k^-8i^cos6x+8z+10 ct
B 1c6k^+8i^cos6x+8z-10 ct
C 1c6k^+8i^cos6x-8z+10 ct
D 1c6k^-8i^cos6x+8z-10 ct

The correct answer is:
1c6k^-8i^cos6x+8z-10 ct

Solution:
E=E0cosωt-K →· r →
∴ At any time t,
K→=6i^+8k^
B0=E0c=10c
λ=2πk=2π10=π5
Also, E→×B→=c→
∴ω=2πf=2πcλ
i^j^k^010BxByBz=6i^+8k^10
ω=2π×cπ5=10c
∴i^Bz-0+j^0+k^-Bx=0.6i^-0.8k^
Comparing the coefficients on both sides,
∴Bz=0.6; Bx=-0.8
B0→=10c-0.8i^+0.6k^=-8i^+6k^c
So the magnetic field,B→=1c-8i^+6k^cos⁡6x+8z-10ct tesla

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The electric field of an electromagnetic wave in free space is given by
$\vec{E}=10 \cos \left(10^7 t+k x\right) \hat{j} V / m$, where $t$ and $x$ are in seconds and metres respectively. It can be inferred that
(1) The wavelength $\lambda$ is $188.4 \mathrm{~m}$.
(2) The wave number $k$ is $0.33 \mathrm{rad} / \mathrm{m}$.
(3) The wave amplitude is $10 \mathrm{~V} / \mathrm{m}$.
(4) The wave is propagating along.t $x$ direction
Which one of the following pairs of statements is correct?

Options:
A (3) and (4)
B (1) and (2)
C (2) and (3)
D (1) and (3)

The correct answer is:
(1) and (3)

Solution:
The electric field of electromagnetic wave
$$
\begin{aligned}
& \overrightarrow{\mathrm{E}}=10 \cos \left(10^7 \mathrm{t} \pm \mathrm{kx}\right) \hat{\mathrm{j}} \\
& \text { Amplitude }=10 \mathrm{~V} / \mathrm{m} \\
& \because \quad \mathrm{c}=\frac{\omega}{\mathrm{k}} \\
& \therefore \quad 3 \times 10^8=\frac{10^7}{\mathrm{k}} \\
& \text { or } \quad k=\frac{1}{30} \\
& \text { or } \quad \frac{2 \pi}{\lambda}=\frac{1}{30} \\
& \text { or } \quad \lambda=188.4 \mathrm{~m} \\
&
\end{aligned}
$$
So, (1) and (3) option are correct.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2010 (Mains)
The electric field of an electromagnetic wave in free space is represented as E→=E0cos(ωt-kz)i^. The corresponding magnetic induction vector will be :

Options:
A B→=E0Ccos(ωt-kz)j^
B B→=E0Ccos(ωt-kz)j^
C B→=E0Ccos(ωt+kz)j^
D B→=E0Ccos(ωt+kz)j^

The correct answer is:
B→=E0Ccos(ωt-kz)j^

Solution:
Given E→=E0cos(ωt-kz)i^:

Clearly, (ωt-kz) term represents, electromagnetic wave is travelling along +z direction or along k^.

Now, in vector form, direction of electromagnetive wave is given by C^=E^×B^. Therefore, magnetic field is along j^.

Also, we know that B0=E0C, where C is speed of light.

Therefore, B→=E0Ccos(ωt-kz)j^.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The electric field of light wave is given as E →=10–3cos⁡2πx5×10–7-2π×6×1014 tx^NC . This light falls on a metal plate of work function 2 eV . The stopping potential of the photo-electrons is:
Given, E (in eV ) =12375λ(in Å)

Options:
A 0.72 V
B 2.0 V
C 2.48 V
D 0.48 V

The correct answer is:
0.48 V

Solution:
E→=10-3cos2πx5×10-7-2π×6×1014tx^ N/c
k=2π5×10-7
And ω=6×1014×2π ⇒f=6×1014 Hz
E=ϕ+KEmax
KEmax=hf-ϕ
=6.6×10-34×6×10141.6×10-19-2
=2.475-2=0.48 eV
∴ Stopping potential is 0.48 V

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The electric field on two sides of a large charged plate is shown in the diagram. The charge density on the plate in S.I. units is given by ( Îµ 0 is the permittivity of free space in S.I. units)

Options:
A 2 ε 0
B 4 Îµ 0
C 10 ε 0
D zero

The correct answer is:
4 ε 0

Solution:
From the figure, it is clear that the plate is placed in an external electric field. Let the electric field due to plate be E, and E0 is the external electric field.

â´ E 0 + E = 1 2 V m - 1
E 0 - E = 8 V m - 1
Solving equations, E = 2 V m-1 and E0 = 10 V m-1
Now, electric field due to plate = Ï 2 Îµ 0 = 2
â´ Ï = 4 Îµ 0

Topic: Physics, Magnetic Properties of Matter, JEE Main
The electric field part of an electromagnetic wave in a medium is represented by
\(\begin{aligned} & E_x=0 \\ & E_y=2.5 \mathrm{NC}^{-1} \cos \left[\left(2 \pi \times 10^6 \mathrm{rad} \mathrm{s}^{-1}\right) \mathrm{t}-\left(\pi \times 10^{-2} \mathrm{rad} \mathrm{m}^{-1}\right) \mathrm{x}\right] \\ & E_z=0\end{aligned}\)
The wave is

Options:
A moving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$.
B moving along $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $100 \mathrm{~m}$
C moving along $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
D moving along $-x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$

The correct answer is:
moving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$.

Solution:
Comparing the given equation
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{y}}=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times 10^6 \frac{\mathrm{rad}}{\mathrm{m}}\right) \mathrm{t}\right. \\
&\left.-\left(\pi \times 10^{-2} \frac{\mathrm{rad}}{\mathrm{sec}}\right) \mathrm{x}\right]
\end{aligned}
$$
With the standard equation
$$
\mathrm{E}_{\mathrm{y}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kx})
$$
we get
$$
\begin{array}{ll}
& \omega=2 \pi \mathrm{f}=2 \pi \times 10^6 \\
\therefore & \mathrm{f}=10^6 \mathrm{~Hz}
\end{array}
$$
Moreover, we know that
$$
\Rightarrow \quad \begin{aligned}
\frac{2 \pi}{\lambda} & =\mathrm{k}=\pi \times 10^{-2} \mathrm{~m}^{-1} \\
\Rightarrow \quad \lambda & =200 \mathrm{~m}
\end{aligned}
$$
Hence, the wave is moving along positive $x$-direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$.

Topic: Physics, Electromagnetic Waves, JEE Main
The electric field part of an electromagnetic wave in a medium is represented by :
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{x}}=0 \\
& \mathrm{E}_{\mathrm{y}}=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times 10^6 \frac{\mathrm{rad}}{\mathrm{m}}\right) \mathrm{t}-\left(\pi \times 10^{-2} \frac{\mathrm{rad}}{\mathrm{s}}\right) \mathrm{x}\right]
\end{aligned}
$$
$E_z=0$. The wave is :

Options:
A Moving along - $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
B Moving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
C Moving along $x$ direction with frequency $10^6$ $\mathrm{Hz}$ and wavelength $100 \mathrm{~m}$
D Moving along $x$ direction with frequency $10^6$ $\mathrm{Hz}$ and wavelength $200 \mathrm{~m}$

The correct answer is:
Moving along $x$ direction with frequency $10^6$ $\mathrm{Hz}$ and wavelength $200 \mathrm{~m}$

Solution:
As the coefficient of $x$ is negative, it is moving along +ve $x$-axis and equating the equation
$$
\begin{aligned}
\mathrm{E}_{\mathrm{y}} & =2.5 \cos \left[\left(2 \pi \times 10^6\right) \mathrm{t}-\left(\pi \times 10^{-2}\right) \mathrm{x}\right] \\
\text { with } \mathrm{y} & =\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{kx}) \\
\omega & =2 \pi \times 106 \\
\Rightarrow \mathrm{f} & =\frac{\omega}{2 \pi}=10^6 \mathrm{~Hz} \\
\mathrm{k} & =\pi \times 10^{-2} \\
\Rightarrow \quad \lambda & =\frac{2 \pi}{\mathrm{k}} \\
& =\frac{2 \pi}{\pi \times 10^{-2}}=200 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves, JEE Main
The electric field part of an electromagnetic wave in a medium is represented by :
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{x}}=0 \\
& \mathrm{E}_{\mathrm{y}}=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times 10^6 \frac{\mathrm{rad}}{\mathrm{m}}\right) \mathrm{t}-\left(\pi \times 10^{-2} \frac{\mathrm{rad}}{\mathrm{s}}\right) \mathrm{x}\right]
\end{aligned}
$$
$E_z=0$. The wave is :

Options:
A Moving along - $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
B Moving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
C Moving along $x$ direction with frequency $10^6$ $\mathrm{Hz}$ and wavelength $100 \mathrm{~m}$
D Moving along $x$ direction with frequency $10^6$ $\mathrm{Hz}$ and wavelength $200 \mathrm{~m}$

The correct answer is:
Moving along $x$ direction with frequency $10^6$ $\mathrm{Hz}$ and wavelength $200 \mathrm{~m}$

Solution:
As the coefficient of $x$ is negative, it is moving along +ve $x$-axis and equating the equation
$$
\begin{aligned}
\mathrm{E}_{\mathrm{y}} & =2.5 \cos \left[\left(2 \pi \times 10^6\right) \mathrm{t}-\left(\pi \times 10^{-2}\right) \mathrm{x}\right] \\
\text { with } \mathrm{y} & =\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{kx}) \\
\omega & =2 \pi \times 106 \\
\Rightarrow \mathrm{f} & =\frac{\omega}{2 \pi}=10^6 \mathrm{~Hz} \\
\mathrm{k} & =\pi \times 10^{-2} \\
\Rightarrow \quad \lambda & =\frac{2 \pi}{\mathrm{k}} \\
& =\frac{2 \pi}{\pi \times 10^{-2}}=200 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves, NEET, NEET 2009 (Mains)
The electric field part of an electromagnetic wave in vacuum is
\( E=3.1 \mathrm{~N} \mathrm{C}^{-1} \cos \left[\left(1.8 \mathrm{rad} \mathrm{m}^{-1}\right) y+\left(5.4 \times 10^{8} \mathrm{rad} \mathrm{s}^{-1}\right) t\right] \hat{\mathrm{i}} \)
The wavelength of this part of electromagnetic wave is,

Options:
A \( 1.5 \mathrm{~m} \)
B \( 2 \mathrm{~m} \)
C \( 2.5 \mathrm{~m} \)
D \( 3.5 \mathrm{~m} \)

The correct answer is:
\( 3.5 \mathrm{~m} \)

Solution:
Given

E=3.1 NC-1×cos1.8 rad m-1y+5.4×108 rad s-1ti^ ...1

Comparing 1 with the equation

E=E0 cosky+ωt ...2

We get, k=1.8, rad m-1

E0=3.1 NC-1, c=3×108 ms-1, ω=5.4×108 rad s-1

Now, λ=2πk=2×221.8×7=3.5 m

Topic: Physics, Electromagnetic Waves, JEE Main
The electric field vector at point P (a, a, a) due to three uniformly charged infinite wires 1, 2 and 3 kept the x, y and z-axes, respectively as shown in shown in figure is (Charge unit length of each wire is λ )

Options:
A λ 3 π ε 0 a i ^ + j ^ + k ^
B λ 2 π ε 0 a i ^ + j ^ + k ^
C λ 2 2 π ε 0 a i &Hat; + j &Hat; + k &Hat;
D 2 λ π ε 0 a i ^ + j ^ + k ^

The correct answer is:
λ 2 π ε 0 a i ^ + j ^ + k ^

Solution:
The electric field vector at
due to (2) E → 2 = k λ a i ^ + k ^
due to (3) E → 3 = k λ a i ^ + j ^
(1) E → 1 = k λ a j ^ + k ^
So E → P = 2 k λ a i ^ + j ^ + k ^

Topic: Physics, Electrostatics, JEE Main
The electric fields of two plane electromagnetic plane waves in vacuum are given by E1→=E0j^cos⁡ωt-kx and E2→=E0k^cos⁡ωt-ky At t=0, a particle of charge q is at origin with a velocity v→=08cj^ ( c is the speed of light in vaccum). The instantaneous force experienced by the particle is:

Options:
A E0q0.8i^-j^+0.4k^
B E0q0.4i^-3j^+0.8k^
C E0q-0.8i^+j^+k^
D E0q0.8i^+j^+0.2k^

The correct answer is:
E0q0.8i^+j^+0.2k^

Solution:
Magnetic field vectors associated with this electromagnetic wave are given by
B→1=E0ck^cos⁡kx-ωt & B→2=E0ci^cos⁡ky-ωt
F→=qE→+qV→×B→
=qE→1+E→2+qV→×B→1+B→2
By putting the value of E→1,E→2,B→1&B→2
The net Lorentz force on the charged particles is
F→=qE00.8cos⁡kx-ωti^+cosky-ωtj^+0.2cosky-ωtk^
At t=0 and at x=y=0
F→=qE00.8i^+j^+0.2k^

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The electric flux for Gaussian surface \( A \) that encloses the charged particles in free space. (Given,
\[
\left.q_{1}=-14 \mathrm{nC}, q_{2}=78.85 \mathrm{nC}, q_{3}=-56 \mathrm{nC} .\right)
\]

Options:
A \( 10^{3} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1} \)
B \( 10^{3} \mathrm{C} \mathrm{N}^{-1} \mathrm{~m}^{-2} \)
C \( 6.32 \times 10^{3} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1} \)
D \( 6.32 \times 10^{3} \mathrm{C} \mathrm{N}^{-1} \mathrm{~m}^{-2} \)

The correct answer is:
\( 10^{3} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1} \)

Solution:
From Gauss's Law we know that,

ϕ=qinε0

where, qin is the net charge enclosed

So, for the given surface net charge enclosed can be calculated,

The total charge enclosed in the surface A is,

qin=q1+q2+q3=-14+78.85-56 nC⇒qin=8.85×10-9 C

Total flux through the surface A is,

ϕ=qinε0=8.85×10-98.85×10-12=103 N m2 C-1

Therefore electric flux for the gaussian surface encloses the charged particle is free space is 103 N m2 C-1.

Topic: Physics, Electrostatics, JEE Main
The electric force between two point charges separated by a certain distance in air is $F$. The distance at which they should be placed in a
medium of relative permittivity $k$, so that the force remains the same, is

Options:
A $d$
B $\frac{d}{k}$
C $k d$
D $\frac{d}{\sqrt{k}}$

The correct answer is:
$\frac{d}{\sqrt{k}}$

Solution:
Let $d$ be the distance in air between two point charges and $d_{m}$ be the distance between them in medium.
According to question,
$F=F_{m}$
$\Rightarrow \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{d^{2}}=\frac{1}{4 \pi \varepsilon_{0} k} \frac{q_{1} q_{2}}{d_{m}^{2}}$
$\Rightarrow \quad d^{2}=k d_{m}^{2}$
$\Rightarrow \quad d_{m}=\frac{d}{\sqrt{k}}$

Topic: Physics, Electrostatics, COMEDK, COMEDK 2012
The electric intensity $E$, current density $j$ and specific resistance $k$ are related to each other by the relation

Options:
A $E=j / k$
B $E=j k$
C $E-k / j$
D $k=j E$

The correct answer is:
$E=j k$

Solution:
We know that $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ and $\mathrm{E}=\frac{\mathrm{I}}{\sigma \mathrm{A}}$; $\therefore \mathrm{E}=\frac{\mathrm{J}}{\sigma}=\rho \mathrm{J}=\mathrm{kJ}$.
hence correct option is option B.

Topic: Physics, Current Electricity, JEE Main
The electric potential at a point in free space due to a charge $\mathrm{Q}$ coulomb is $\mathrm{Q} \times 10^{11}$ volts. The electric field at that point is

Options:
A $12 \pi \varepsilon_0 Q \times 10^{22}$ volt/m
B $4 \pi \varepsilon_0 Q \times 10^{22}$ volt/m
C $12 \pi \varepsilon_0 \mathrm{Q} \times 10^{20} \mathrm{volt} / \mathrm{m}$
D $4 \pi \varepsilon_0 \mathrm{Q} \times 10^{20}$ volt/m

The correct answer is:
$4 \pi \varepsilon_0 Q \times 10^{22}$ volt/m

Solution:
$$
\begin{aligned}
V & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R}=Q \times 10^{11} \text { volt } \\
E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R^2}=\frac{V}{R}=Q \times 10^{11} \times 4 \pi \varepsilon_0 \times 10^{11}[\text { from ...(i) }] \\
& =4 \pi \varepsilon_0 Q \times 10^{22} \text { volt } / \mathrm{m}
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2008 (Mains)
The electric potential at a point on the axis of an electric dipole depends on the distance $r$ of the point from the dipole as

Options:
A $\propto \frac{1}{r}$
B $\propto \frac{1}{r^2}$
C $\propto r$
D $\propto \frac{1}{r^3}$

The correct answer is:
$\propto \frac{1}{r^2}$

Solution:
Electric potential due to dipole in it's general position is given by $V=\frac{k \cdot p \cos \theta}{r^2} \Rightarrow V \propto \frac{1}{r^2}$

Topic: Physics, Electrostatics, JEE Main
The electric potential at a point x,y in the x-y plane is given by

V=-Kxy

The electric field intensity at a distance r from the origin varies as

Options:
A r2
B r
C 2r
D 2r2

The correct answer is:
r

Solution:
Co-ordinates of the point are (x, y)
Distance of point from origin,
r=x2+y2, V=-kxy
Ex=-∂V∂x=-∂∂x-kxy=ky
Ey=-∂V∂y=-∂∂y-kxy=kx
E=Ex2+Ey2=ky2+x2=kr

Topic: Physics, Electrostatics, JEE Main
The electric potential at a point $(x, y, z)$ is given by $\mathrm{V}=-x^2 y-x z^3+4$
The electric field $\overrightarrow{\mathrm{E}}$ at that point is :

Options:
A $\vec{E}=\hat{i}\left(2 x y-z^3\right)+\hat{j} x y^2+\hat{k} 3 z^2 x$
B $\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}}\left(2 x y+\mathrm{z}^3\right)+\hat{\mathrm{j}} \mathrm{x}^2+\hat{\mathrm{k}} 3 \mathrm{xz} \mathrm{z}^2$
C $\vec{E}=\hat{i} 2 x y+\hat{j}\left(x^2+y^2\right)+\hat{k}\left(3 x z-y^2\right)$
D $\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} z+\hat{\mathrm{j}} x y z+\hat{\mathrm{k}} z^2$

The correct answer is:
$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}}\left(2 x y+\mathrm{z}^3\right)+\hat{\mathrm{j}} \mathrm{x}^2+\hat{\mathrm{k}} 3 \mathrm{xz} \mathrm{z}^2$

Solution:
$$
\begin{aligned}
V & =-x^2 y-x z^3+4 \\
\vec{E} & =-V=-\left(\hat{i} \frac{\delta}{\delta x}+\hat{j} \frac{\delta}{\delta y}+\hat{k} \frac{\delta}{\delta z}\right) \\
& \left.=\left(2 x y+z^3\right) \hat{i}+x^2 \hat{j}+3 x z^2+4\right)
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2009 (Mains)
The electric potential at a space point P x, y, z is given as V=x2+y2+z2. The modulus of the electric field at that point is proportional to

Options:
A V12
B V
C V32
D V2

The correct answer is:
V12

Solution:
Ex=-2x, Ey=-2 and Ez=-2z
E=Ex2+Ey2+Ez2
=4x2+4y2+4z2
=2x2+y2+z2=2V

Topic: Physics, Electrostatics, JEE Main
The electric potential at the centre of two concentric half rings of radii R1 and R2, having same linear charge density λ is

.

Options:
A 2λ∈0
B λ2∈0
C λ4∈0
D λ∈0

The correct answer is:
λ2∈0

Solution:
VO=V1+V2

⇒VO=14πϵ0×λ1πR1R1+14πϵ0×λ1πR2R2=λ2ϵ0

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The electric potential at the surface of an atomic nucleus (Z=50) of radius 9×10-13 cm is α×106 V. What is the value of α?

(Charge of proton 1.6×10-19 C)

Options:

The correct answer is:
8

Solution:
The electric potential at the surface of an atomic nucleus can be written as

V=kQR=k·ZeR ...1

From equation (1), it follows that

V=9×109×50×1.6×10-199×10-13×10-2 V=8×106 V

Hence, α=8.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The electric potential in a region is given as V=â4ar2+3b, where r is distance from the origin, a and b are constants. If the volume charge density in the region is given by Ï=naÎµ0, then what is the value of n?

Options:

The correct answer is:
24

Solution:
V=-4ar2+3b

E=-dV dr=8ar

The small amount of flux dϕ associated with the closed surface bounded by two spheres of radius r and r+dr is

dϕ=E+dEA+dA-EA

dϕ=EdA+dEA

dE=8adr and dA=8πrdr

Using Gauss's law

ρ×4πr2drε0=8adr×4πr2+8ar×8πrdr

⇒ ρ = 24aε0

n = 24

Topic: Physics, Electrostatics, JEE Main