2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
Two charged thin infinite plane sheets of
uniform surface charge density ${\sigma _ + }$ and ${\sigma _ - }$,
where |${\sigma _ + }$| > |${\sigma _ - }$|, intersect at right angle.
Which of the following best represents the
electric field lines for this system :
uniform surface charge density ${\sigma _ + }$ and ${\sigma _ - }$,
where |${\sigma _ + }$| > |${\sigma _ - }$|, intersect at right angle.
Which of the following best represents the
electric field lines for this system :
A.
B.
C.
D.
Given, |${\sigma _ + }$| > |${\sigma _ - }$|
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
A two point charges 4q and -q are fixed
on the x-axis at x = $ - {d \over 2}$
and x = ${d \over 2}$
respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will :
A.
increase by ${{3{q^2}} \over {4\pi {\varepsilon _0}d}}$
B.
increase by ${{2{q^2}} \over {3\pi {\varepsilon _0}d}}$
C.
decrease by ${{{q^2}} \over {4\pi {\varepsilon _0}d}}$
D.
decrease by ${{4{q^2}} \over {3\pi {\varepsilon _0}d}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
Concentric metallic hollow spheres of radii R and 4R hold charges Q1
and Q2
respectively. Given that
surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is :
A.
${{3{Q_2}} \over {4\pi {\varepsilon _0}R}}$
B.
${{3{Q_1}} \over {4\pi {\varepsilon _0}R}}$
C.
${{3{Q_1}} \over {16\pi {\varepsilon _0}R}}$
D.
${{{Q_2}} \over {4\pi {\varepsilon _0}R}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
Two isolated conducting spheres S1 and S2 of
radius ${2 \over 3}R$ and ${1 \over 3}R$ have 12 $\mu $C and –3 $\mu $C
charges, respectively, and are at a large
distance from each other. They are now
connected by a conducting wire. A long time
after this is done the charges on S1 and S2 are
respectively :
A.
4.5 $\mu $C on both
B.
+4.5 $\mu $C and –4.5 $\mu $C
C.
6 $\mu $C and 3 $\mu $C
D.
3 $\mu $C and 6 $\mu $C
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
A small point mass carrying some positive
charge on it, is released from the edge of a
table. There is a uniform electric field in this
region in the horizontal direction. Which of the
following options then correctly describe the
trajectory of the mass?
(Curves are drawn schematically and are not to scale).
(Curves are drawn schematically and are not to scale).
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
A charge Q is distributed over two concentric
conducting thin spherical shells radii r and
R (R > r). If the surface charge densities on the
two shells are equal, the electric potential at
the common centre is :
A.
${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}$Q
B.
${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {2\left( {{R^2} + {r^2}} \right)}}Q$
C.
${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + 2r} \right)Q} \over {2\left( {{R^2} + {r^2}} \right)}}$
D.
${1 \over {4\pi {\varepsilon _0}}}{{\left( {2R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}Q$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
A charged particle (mass m and charge q)
moves along X-axis with velocity V0. When it
passes through the origin it enters a region having uniform electric field
$\overrightarrow E = - E\widehat j$ which extends upto x = d.
Equation of path of electron in the region x > d is
moves along X-axis with velocity V0. When it
passes through the origin it enters a region having uniform electric field
$\overrightarrow E = - E\widehat j$ which extends upto x = d.
Equation of path of electron in the region x > d is
A.
y = ${{qEd} \over {mV_0^2}}\left( {x - d} \right)$
B.
y = ${{qEd} \over {mV_0^2}}\left( {{d \over 2} - x} \right)$
C.
y = ${{qEd} \over {mV_0^2}}x$
D.
y = ${{qE{d^2}} \over {mV_0^2}}x$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
Consider a sphere of radius R which carries a
uniform charge density $\rho $. If a sphere of radius ${{R \over 2}}$ is carved out of it, as shown, the ratio ${{\left| {\overrightarrow {{E_A}} } \right|} \over {\left| {\overrightarrow {{E_B}} } \right|}}$ of magnitude of electric field ${\overrightarrow {{E_A}} }$ and ${\overrightarrow {{E_B}} }$,
respectively, at points A and B due to the
remaining portion is :
A.
${{17} \over {54}}$
B.
${{18} \over {54}}$
C.
${{18} \over {34}}$
D.
${{21} \over {34}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
An electric dipole of moment
$\overrightarrow p = \left( { - \widehat i - 3\widehat j + 2\widehat k} \right) \times {10^{ - 29}} $ C.m is
at the origin (0, 0, 0). The electric field due to this dipole at
$\overrightarrow r = + \widehat i + 3\widehat j + 5\widehat k$ (note that $\overrightarrow r .\overrightarrow p = 0$ ) is parallel to :
$\overrightarrow p = \left( { - \widehat i - 3\widehat j + 2\widehat k} \right) \times {10^{ - 29}} $ C.m is
at the origin (0, 0, 0). The electric field due to this dipole at
$\overrightarrow r = + \widehat i + 3\widehat j + 5\widehat k$ (note that $\overrightarrow r .\overrightarrow p = 0$ ) is parallel to :
A.
$\left( { + \widehat i + 3\widehat j - 2\widehat k} \right)$
B.
$\left( { + \widehat i - 3\widehat j - 2\widehat k} \right)$
C.
$\left( { - \widehat i + 3\widehat j - 2\widehat k} \right)$
D.
$\left( { - \widehat i - 3\widehat j + 2\widehat k} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
Consider two charged metallic spheres S1 and
S2 of radii R1 and R2, respectively. The electric
fields E1 (on S1) and E2 (on S2) on their surfaces
are such that E1/E2 = R1/R2. Then the ratio
V1 (on S1) / V2 (on S2) of the electrostatic
potentials on each sphere is :
A.
(R1/R2)2
B.
(R2/R1)
C.
(R1/R2)3
D.
R1/R2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
A particle of mass m and charge q is released
from rest in a uniform electric field. If there is
no other force on the particle, the dependence
of its speed v on the distance x travelled by it
is correctly given by (graphs are schematic and
not drawn to scale)
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
In finding the electric field using Gauss Law
the formula $\left| {\overrightarrow E } \right| = {{{q_{enc}}} \over {{\varepsilon _0}\left| A \right|}}$ is applicable. In the
formula ${{\varepsilon _0}}$ is permittivity of free space, A is the
area of Gaussian surface and qenc is charge
enclosed by the Gaussian surface. The equation
can be used in which of the following situation?
A.
Only when $\left| {\overrightarrow E } \right|$ = constant on the surface.
B.
For any choice of Gaussian surface.
C.
Only when the Gaussian surface is an
equipotential surface.
D.
Only when the Gaussian surface is an
equipotential surface and $\left| {\overrightarrow E } \right|$ is constant on
the surface.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
Three charged particle A, B and C with charges
–4q, 2q and –2q are present on the
circumference of a circle of radius d. the charged
particles A, C and centre O of the circle formed
an equilateral triangle as shown in figure. Electric
field at O along x-direction is :
A.
${3{\sqrt 3 q} \over 4{\pi {\varepsilon _0}{d^2}}}$
B.
${{\sqrt 3 q} \over 4{\pi {\varepsilon _0}{d^2}}}$
C.
${{\sqrt 3 q} \over {\pi {\varepsilon _0}{d^2}}}$
D.
${{2\sqrt 3 q} \over {\pi {\varepsilon _0}{d^2}}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
Two infinite planes each with uniform surface charge density to are kept in such a way that the angle between them is 30o. The electric field in the region shown between them is given by :
A.
${\sigma \over {{ \in _0}}}\left[ {\left( {1 + {{\sqrt 3 } \over 2}} \right)\widehat y + {{\widehat x} \over 2}} \right]$
B.
${\sigma \over {2{ \in _0}}}\left[ {\left( {1 + \sqrt 3 } \right)\widehat y + {{\widehat x} \over 2}} \right]$
C.
${\sigma \over {2{ \in _0}}}\left[ {\left( {1 + \sqrt 3 } \right)\widehat y - {{\widehat x} \over 2}} \right]$
D.
${\sigma \over {2{ \in _0}}}\left[ {\left( {1 - {{\sqrt 3 } \over 2}} \right)\widehat y - {{\widehat x} \over 2}} \right]$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
An electric field $\overrightarrow E = 4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$ N/C
passes through the box shown in figure. The
flux of the electric field through surfaces ABCD
and BCGF are marked as ${\phi _I}$ and ${\phi _{II}}$
respectively. The difference between $\left( {{\phi _I} - {\phi _{II}}} \right)$ is (in Nm2/C) _______.
passes through the box shown in figure. The
flux of the electric field through surfaces ABCD
and BCGF are marked as ${\phi _I}$ and ${\phi _{II}}$
respectively. The difference between $\left( {{\phi _I} - {\phi _{II}}} \right)$ is (in Nm2/C) _______.
Correct Answer: -48
Explanation:
Flux via ABCD
$\phi $I = $\int {\overrightarrow E } .d\overrightarrow A $ = 0
Flux via EFGH
$\phi $II = $\int {\overrightarrow E } .d\overrightarrow A $
= [$4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$].4$\widehat i$
= 16x = 16 $ \times $ 3 = 48
${\phi _I} - {\phi _{II}}$ = 0 - 48 = -48 Nm2/C
$\phi $I = $\int {\overrightarrow E } .d\overrightarrow A $ = 0
Flux via EFGH
$\phi $II = $\int {\overrightarrow E } .d\overrightarrow A $
= [$4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$].4$\widehat i$
= 16x = 16 $ \times $ 3 = 48
${\phi _I} - {\phi _{II}}$ = 0 - 48 = -48 Nm2/C
2020
JEE Advanced
MSQ
JEE Advanced 2020 Paper 2 Offline
Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is $\alpha$. The spheres are now immersed in a dielectric liquid of density 800 kg m$-$3 and dielectric constant 21. If the angle between the strings remains the same after the immersion, then
A.
electric force between the spheres remains unchanged
B.
electric force between the spheres reduces
C.
mass density of the spheres is 840 kg m$-$3
D.
the tension in the strings holding the spheres remains unchanged
2020
JEE Advanced
MSQ
JEE Advanced 2020 Paper 1 Offline
A uniform electric field, $\overrightarrow E = - 400\sqrt 3 \widehat y$ NC−1
is applied in a region. A charged particle of mass m
carrying positive charge q is projected in this region with an initial speed of 2$\sqrt {10} $ $ \times $ 106 ms−1
. This
particle is aimed to hit a target T, which is 5 m away from its entry point into the field as shown
schematically in the figure.
Take ${q \over m}$ = 1010 Ckg−1 . Then
Take ${q \over m}$ = 1010 Ckg−1 . Then
A.
the particle will hit T if projected at an angle 45o
from the horizontal
B.
the particle will hit T if projected either at an angle 30o or 60o
from the horizontal
C.
time taken by the particle to hit T could be $\sqrt {{5 \over 6}} $ $\mu $s as well as $\sqrt {{5 \over 2}} $ $\mu $s
D.
time taken by the particle to hit T is $\sqrt {{5 \over 3}} $ $\mu $s
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole
moment $\overrightarrow p $ is now brought towards q from infinity so that the charge moves away. The final
equilibrium position of the system including the direction of the dipole, the angles and distances is
shown in the figure below. If the work done in bringing the dipole to this position is N $ \times $ (mgh),
where g is the acceleration due to gravity, then the value of N is _________ .
(Note that for three coplanar forces keeping a point mass in equilibrium, ${F \over {\sin \theta }}$ is the same for all forces, where F is any one of the forces and $\theta $ is the angle between the other two forces)
(Note that for three coplanar forces keeping a point mass in equilibrium, ${F \over {\sin \theta }}$ is the same for all forces, where F is any one of the forces and $\theta $ is the angle between the other two forces)
Correct Answer: 2
Explanation:
Ui = 0
${U_f} = {{kqp} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}}$ + mgh .... (i)
Now, from $\Delta$OAB,
$\alpha$ + 90 $-$ $\theta$ + 90 $-$ $\theta$ = 180
$\Rightarrow$ $\alpha$ = 2$\theta$
From $\Delta$ABC, h = 2l sin$\left( {{\alpha \over 2}} \right)$ sin$\theta$
h = 2l sin$\left( {{\alpha \over 2}} \right)$ sin$\left( {{\alpha \over 2}} \right)$ $\Rightarrow$ h = 2l sin2$\left( {{\alpha \over 2}} \right)$
Now, charge is in equilibrium at point B.
So, using sine rule,
${{mg} \over {\sin \left( {90 + {\alpha \over 2}} \right)}} = {{qE} \over {\sin 1(80 - 2\theta )}}$
$ \Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {\sin 2\theta }}$
$ \Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {\sin \alpha }} \Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {2\sin {\alpha \over 2}\cos {\alpha \over 2}}}$
$ \Rightarrow qE = mg2\sin \left( {{\alpha \over 2}} \right)$
$ \Rightarrow {{q2kp} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^3}}} = mg2\sin \left( {{\alpha \over 2}} \right)$
$ \Rightarrow {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} = mg\sin \left( {{\alpha \over 2}} \right) \times \left( {2l\sin {\alpha \over 2}} \right)$
$ \Rightarrow {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} = mgh$
Substituting this in Eq. (i), we get
${U_f} = mgh + {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} \Rightarrow {U_f} = 2mgh$
W = $\Delta$U = 2mgh = Nmgh
$ \therefore $ N = 2
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as
shown in the figure. Charged oil drops of density 900 kg m−3
are released through a tiny hole at the
center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a
voltage of 200 V across the discs. As a result, an oil drop of radius 8 $ \times $ 10−7 m stops moving
vertically and floats between the discs. The number of electrons present in this oil drop is ________.
(neglect the buoyancy force, take acceleration due to gravity = 10 ms−2 and charge on an electron (e) = 1.6 $ \times $ 10–19 C)
(neglect the buoyancy force, take acceleration due to gravity = 10 ms−2 and charge on an electron (e) = 1.6 $ \times $ 10–19 C)
Correct Answer: 6
Explanation:
$qE = Mg $
$\Rightarrow q = {{mg} \over E} = \left( {{{mg} \over {V/d}}} \right)$
$ = {{900 \times {4 \over 5}\pi (8 \times {{10}^{ - 7}}) \times 10} \over {{{200} \over {0.01}}}}$
$N = {q \over e} = {{900 \times 4\pi \times {8^3} \times {{10}^{ - 21}} \times 10 \times 0.01} \over {200 \times 1.6 \times {{10}^{ - 19}}}}$
N = 6
$\Rightarrow q = {{mg} \over E} = \left( {{{mg} \over {V/d}}} \right)$
$ = {{900 \times {4 \over 5}\pi (8 \times {{10}^{ - 7}}) \times 10} \over {{{200} \over {0.01}}}}$
$N = {q \over e} = {{900 \times 4\pi \times {8^3} \times {{10}^{ - 21}} \times 10 \times 0.01} \over {200 \times 1.6 \times {{10}^{ - 19}}}}$
N = 6
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 1 Offline
A circular disc of radius R carries surface charge density
$\sigma \left( r \right) = {\sigma _0}\left( {1 - {r \over R}} \right)$, where $\sigma $0 is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi $0. Electric flux through another spherical surface of radius ${R \over 4}$ and concentric with the disc is $\phi $. Then the ratio ${{{\phi _0}} \over \phi }$ is ________.
$\sigma \left( r \right) = {\sigma _0}\left( {1 - {r \over R}} \right)$, where $\sigma $0 is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi $0. Electric flux through another spherical surface of radius ${R \over 4}$ and concentric with the disc is $\phi $. Then the ratio ${{{\phi _0}} \over \phi }$ is ________.
Correct Answer: 6.4
Explanation:
To determine the disc’s total charge, integrate the radial surface–charge distribution across its entire area:
$ \begin{aligned} q_{\text{enc}} &= \int_{0}^{a} \sigma(r)\, 2\pi r\,dr = 2\pi \sigma_{0}\int_{0}^{a}\!\left(r - \frac{r^{2}}{a}\right)dr \\[4pt] &= 2\pi \sigma_{0}\!\left[\frac{r^{2}}{2} - \frac{r^{3}}{3a}\right]_{0}^{a} = \frac{\pi \sigma_{0}a^{2}}{3}. \end{aligned} $
The electric flux through the spherical surface that encloses the entire charged disc is
$ \phi_0=\frac{q_{\mathrm{enc} 0}}{\epsilon_0}=\frac{\pi \sigma_0 a^2}{3 \epsilon_0} . $
Next, we find the flux through the smaller spherical surface of radius $a / 4$, which encloses a portion of the disc. The charge on this portion of the disc is
$ q_{\mathrm{enc}}=2 \pi \sigma_0 \int\limits_0^{a / 4}\left(r-\frac{r^2}{a}\right) \mathrm{d} r=\frac{5}{96} \pi \sigma_0 a^2 $
The flux through this spherical surface is
$ \phi=\frac{q_{\mathrm{enc}}}{\epsilon_0}=\frac{5 \pi \sigma_0 a^2}{96 \epsilon_0} $
The ratio of the two fluxes is
$ \frac{\phi_0}{\phi}=\frac{32}{5}=6.4 $
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 1 Offline
One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0, 0).
A point particle of mass m carrying a positive charge q is attached at its other end. The entire system
is kept on a smooth horizontal surface. When a point dipole $\overrightarrow p $ pointing towards the charge q is fixed
at the origin, the spring gets stretched to a length l and attains a new equilibrium position (see figure
below). If the point mass is now displaced slightly by $\Delta $l << l from its equilibrium position and
released, it is found to oscillate at frequency ${1 \over \delta }\sqrt {{k \over m}} $. The value of $\delta $ is ______.
Correct Answer: 3.14
Explanation:
Electrostatic force = Spring force
At l, $Fe = {F_{sp}}kl = {{2\alpha pq} \over {{l^3}}}$ (Here, $\alpha = {1 \over {4\pi {\varepsilon _0}}}$)
Now, the mass m is displaced by $\Delta $l = x from the mean position.
${F_{net}} = {F_{sp}} - Fe = k(l + x) - {{q(2\alpha p)} \over {{{(l + x)}^3}}}$
$ = k(x + 1) - {{q(2\alpha p)} \over {{l^3}{{(l + x/l)}^3}}}$
$ = kx + kl - q\left( {{{2\alpha p} \over {{l^3}}}} \right)\left( {1 - {{3x} \over l}} \right)$
$ = kx + kl - q\left( {{{2\alpha p} \over {{l^3}}}} \right) + {{2\alpha p} \over {{l^3}}}.{{3x} \over l}$
Substituting ${{2\alpha pq} \over {{l^3}}} = kl$, we get
${F_{net}} = kx + kl\left( {{{3x} \over l}} \right) = 4kx$
This is restoring in nature.
Hence, ${k_{eq}} = 4k$
or $T = 2\pi \sqrt {{m \over {4k}}} = \pi \sqrt {{m \over k}} $
$ \therefore $ $f = {1 \over \pi }\sqrt {{k \over m}} $
So, $\delta = \pi = 3.14$
At l, $Fe = {F_{sp}}kl = {{2\alpha pq} \over {{l^3}}}$ (Here, $\alpha = {1 \over {4\pi {\varepsilon _0}}}$)
Now, the mass m is displaced by $\Delta $l = x from the mean position.
${F_{net}} = {F_{sp}} - Fe = k(l + x) - {{q(2\alpha p)} \over {{{(l + x)}^3}}}$
$ = k(x + 1) - {{q(2\alpha p)} \over {{l^3}{{(l + x/l)}^3}}}$
$ = kx + kl - q\left( {{{2\alpha p} \over {{l^3}}}} \right)\left( {1 - {{3x} \over l}} \right)$
$ = kx + kl - q\left( {{{2\alpha p} \over {{l^3}}}} \right) + {{2\alpha p} \over {{l^3}}}.{{3x} \over l}$
Substituting ${{2\alpha pq} \over {{l^3}}} = kl$, we get
${F_{net}} = kx + kl\left( {{{3x} \over l}} \right) = 4kx$
This is restoring in nature.
Hence, ${k_{eq}} = 4k$
or $T = 2\pi \sqrt {{m \over {4k}}} = \pi \sqrt {{m \over k}} $
$ \therefore $ $f = {1 \over \pi }\sqrt {{k \over m}} $
So, $\delta = \pi = 3.14$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
The electric flux from a cube of edge $l$ is $\phi$ in an enclosed charge. If the edge of the cube is made $\frac{2}{3} l$ and the charge enclosed in the cube is doubled, then the electric flux value will be
A.
$4 \phi$
B.
$2 \phi$
C.
$\frac{\phi}{2}$
D.
$\phi$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
- If the dielectric constant of a substance $K=\frac{4}{3}$, then the electric susceptibility $\chi$ in terms of vacuum permittivity $\varepsilon_0$ is
A.
$\frac{\varepsilon_0}{3}$
B.
$3 \varepsilon_0$
C.
$\frac{4}{3} \varepsilon_0$
D.
$\frac{3}{4} \varepsilon_0$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
In a uniformly charged sphere of total charge $Q$ and radius $R$, the electric field $E$ is plotted as function of distance from the centre of the sphere. The graph which would correspond to the above description will be
A.
B.
C.
D.
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
Electric charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively which are at a distance of $2 L$ apart. If $C$ is the midpoint between $A$ and $B$, then the work done in moving a charge $+Q$ along the semi-circle $C R D$ is
A.
$\frac{q Q}{2 \pi \varepsilon_0 L}$
B.
$\frac{q Q}{6 \pi \varepsilon_0 L}$
C.
$-\frac{q Q}{6 \pi \varepsilon_0 L}$
D.
$\frac{q Q}{4 \pi \varepsilon_0 L}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
Two negative charges of equal magnitude are located in $x y$-plane as shown below in the figure. The direction of the electric field at point $P$ is
A.
along positive, $x$-direction
B.
along negative, $x$-direction
C.
along positive, $y$-direction
D.
along negative, $y$-direction
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
An infinite non-conducting sheet has a surface charge density $2 \times 10^{-7} \mathrm{C} / \mathrm{m}^2$ on one side. The distance between two equipotential surfaces whose potential differ by 90 V is (assume, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$ )
A.
$20 \pi \mathrm{~mm}$
B.
$\frac{25}{\pi} \mathrm{~mm}$
C.
$\frac{12.5}{\pi} \mathrm{~mm}$
D.
$\frac{\pi}{20} \mathrm{~mm}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
If a proton is moved against the coulomb force of an electric field, then
A.
work is done by the electric field
B.
energy is used from some outside source
C.
the strength of the field is decreased
D.
the strength of the field is increased
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by $\rho $(r) = kr, where
r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite
points, at equal distance, $a$ from the centre. If A and B do not experience any force, then :
A.
$a = {8^{ - 1/4}}R$
B.
$a = {2^{ - 1/4}}R$
C.
$a = {{3R} \over {{2^{1/4}}}}$
D.
$a = {R \over {\sqrt 3 }}$
Charge enclosed in the sphere,
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
A point dipole $\overrightarrow p = - {p_0}\widehat x$
is kept at the origin. The potential and electric field due to this dipole on the
y-axis at a distance d are, respectively: (Take V= 0 at infinity)
A.
${{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}$
B.
$0,{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}$
C.
${{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}$
D.
$0,{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge
Q. At its centre is a dipole $\overrightarrow P $
as shown. In this case :
A.
surface charge density on the inner surface is uniform and equal to ${{\left( {Q/2} \right)} \over {4\pi {a^2}}}$
B.
surface charge density on the inner surface of the shell is zero everywhere
C.
surface charge density on the outer surface depends on $\left| {\overrightarrow P } \right|$
D.
electric field outside the shell is the same as that of a point charge at the centre of the shell
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
In free space, a particle A of charge 1$\mu $C is held fixed at a point P. Another particle B of the same charge and
mass 4$\mu $g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P
is :
$\left[ {Take\,{1 \over {4\pi { \in _0}}} = 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right]$
A.
1.0 m/s
B.
6.32 $ \times $ 104 m/s
C.
2.0 $ \times $ 103 m/s
D.
1.5 $ \times $ 102 m/s
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
A uniformly charged ring of radius 3a and total
charge q is placed in xy-plane centred at origin.
A point charge q is moving towards the ring
along the z-axis and has speed u at z = 4a. The
minimum value of u such that it crosses the
origin is :
A.
$\sqrt {{2 \over m}} {\left( {{2 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$
B.
$\sqrt {{2 \over m}} {\left( {{1 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$
C.
$\sqrt {{2 \over m}} {\left( {{1 \over {5}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$
D.
$\sqrt {{2 \over m}} {\left( {{4 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
Four point charges –q, +q, +q and –q are placed
on y-axis at y = –2d, y = –d, y = +d and
y = +2d, respectively. The magnitude of the
electric field E at a point on the x-axis at
x = D, with D >> d, will behave as :-
A.
$E \propto {1 \over D^3}$
B.
$E \propto {1 \over D}$
C.
$E \propto {1 \over D^4}$
D.
$E \propto {1 \over D^2}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
A system of three charges are placed as shown
in the figure :
If D >> d, the potential energy of the system is
best given by :
If D >> d, the potential energy of the system is
best given by :
A.
${1 \over {4\pi {\varepsilon _0}}}\left[ { {{{q^2}} \over d} + {{qQd} \over {{D^2}}}} \right]$
B.
${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} - {{qQd} \over {2{D^2}}}} \right]$
C.
${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} - {{qQd} \over {{D^2}}}} \right]$
D.
${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} + {2{qQd} \over {{D^2}}}} \right]$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
A positive point charge is released from rest at
a distance r0 from a positive line charge with
uniform density. The speed (v) of the point
charge, as a function of instantaneous distance
r from line charge, is proportional to :-
A.
$v \propto \left( {{r \over {{r_0}}}} \right)$
B.
$v \propto \ln \left( {{r \over {{r_0}}}} \right)$
C.
$v \propto {e^{ + r/{r_0}}}$
D.
$v \propto \sqrt {\ln \left( {{r \over {{r_0}}}} \right)} $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
The electric field in a region is given by
$\mathop E\limits^ \to = \left( {Ax + B} \right)\mathop i\limits^ \wedge $
, where E is in NC–1 and x is in
metres. The values of constants are
A = 20 SI unit and B = 10 SI unit. If the potential
at x = 1 is V1 and that at x = –5 is V2, then
V1 – V2 is :-
A.
–520 V
B.
180 V
C.
–48 V
D.
320 V
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
The bob of a simple pendulum has mass 2g and
a charge of 5.0 μC. It is at rest in a uniform
horizontal electric field of intensity 2000 V/m.
At equilibrium, the angle that the pendulum
makes with the vertical is : (take g = 10 m/s2)
A.
tan–1(5.0)
B.
tan–1(2.0)
C.
tan–1(0.5)
D.
tan–1(0.2)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
A solid conducting sphere, having a charge Q,
is surrounded by an uncharged conducting
hollow spherical shell. Let the potential
difference between the surface of the solid
sphere and that of the outer surface of the
hollow shell be V. If the shell is now given a
charge of –4 Q, the new potential difference
between the same two surfaces is :
A.
V
B.
2V
C.
–2V
D.
4V
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
Determine the electric dipole moment of the system of the three charges, placed on the vertices of an equilateral triangle, as shown in the figure :
A.
$2q\ell \widehat j$
B.
$\left( {q\ell } \right){{\widehat i + \widehat j} \over {\sqrt 2 }}$
C.
$\sqrt 3 \,q\ell {{\widehat j - \widehat i} \over {\sqrt 2 }}$
D.
$ - \sqrt 3 \,q\ell \widehat j$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R (t) is :
A.
B.
C.
D.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
An electric field of 1000 V/m is applied to an electric dipole at angle of 45o. The value of electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole?
A.
- 7 $ \times $ 10–27 J
B.
$-$ 9 $ \times $ 10–20 J
C.
$-$ 10 $ \times $ 10–29 J
D.
$-$ 20 $ \times $ 10–18 J
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
Three charges Q, + q and + q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is :
A.
${{ - q} \over {1 + \sqrt 2 }}$
B.
+ q
C.
$-$ 2q
D.
${{ - \sqrt 2 q} \over {\sqrt 2 + 1}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
The given graph shows variation (with distance r form centre) of :
A.
Electric field of a uniformly charged sphere
B.
Electric field of a uniformly charged spherical shell
C.
Potential of a uniformly charged sphere
D.
Potential of a uniformly charged spherical shell
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be -
A.
${{{Q_2}} \over {4\pi {\varepsilon _0}}}$
B.
${{{Q^2}} \over {2\sqrt 2 \pi {\varepsilon _0}}}$
C.
${{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 3 }}} \right)$
D.
${{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
Charges –q and +q located at A and B, respectively, constitude an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P' such that OP' = $\left( {{y \over 3}} \right)$, the force on Q will be close to - $\left( {{y \over 3} > > 2a} \right)$
A.
9F
B.
3F
C.
F/3
D.
27F
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be -
A.
${{Q\left( {{a^2} + {b^2} + {c^2}} \right)} \over {4\pi {\varepsilon _0}\left( {{a^3} + {b^3} + {c^3}} \right)}}$
B.
${Q \over {4\pi {\varepsilon _0}\left( {a + b + c} \right)}}$
C.
${Q \over {12\pi {\varepsilon _0}}}{{ab + bc + ca} \over {abc}}$
D.
${{Q\left( {a + b + c} \right)} \over {4\pi {\varepsilon _0}\left( {{a^2} + {b^2} + {c^2}} \right)}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
Two electric dipoles, A, B with respective dipole moments ${\overrightarrow d _A} = - 4qai$ and ${\overrightarrow d _B} = - 2qai$ are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is -
A.
${{\sqrt 2 R} \over {\sqrt 2 + 1}}$
B.
${R \over {\sqrt 2 + 1}}$
C.
${{\sqrt 2 R} \over {\sqrt 2 - 1}}$
D.
${R \over {\sqrt 2 - 1}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
Two point charges q1$\left( {\sqrt {10} \mu C} \right)$ and q2($-$ 25 $\mu $C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
[take ${1 \over {4\pi { \in _0}}}$ = 9 $ \times $ 109 Nm2C$-$2]
[take ${1 \over {4\pi { \in _0}}}$ = 9 $ \times $ 109 Nm2C$-$2]
A.
$\left( {63\widehat i - 27\widehat j} \right) \times {10^2}$
B.
$\left( { - 63\widehat i + 27\widehat j} \right) \times {10^2}$
C.
$\left( {81\widehat i - 81\widehat j} \right) \times {10^2}$
D.
$\left( { - 81\widehat i + 81\widehat j} \right) \times {10^2}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
Charge is distributed within a sphere of radius R with a volume charge density $\rho \left( r \right) = {A \over {{r^2}}}{e^{ - {{2r} \over s}}},$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :
A.
a log $\left( {1 - {Q \over {2\pi aA}}} \right)$
B.
${a \over 2}$ log $\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$
C.
a log $\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$
D.
${a \over 2}$ log $\left( {1 - {Q \over {2\pi aA}}} \right)$
