1
JEE Advanced 2020 Paper 1 Offline
Numerical
+4
-0
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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 ________.
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2
JEE Advanced 2018 Paper 2 Offline
Numerical
+3
-0
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A particle, of mass $${10^{ - 3}}$$ $$kg$$ and charge $$1.0$$ $$C,$$ is initially at rest. At time $$t=0,$$ the particle comes under the influence of an electric field $$\overrightarrow E \left( t \right) = {E_0}\sin \,\,$$ $$\omega t\widehat i,$$ where $${E_0} = 1.0\,N{C^{ - 1}}$$ and $$\omega = 10{}^3\,rad\,{s^{ - 1}}.$$ Consider the effect of only the electrical force on the particle. Then the maximum speed, in $$m{s^{ - 1}},$$ attained by the particle at subsequent times is _______________.
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3
JEE Advanced 2015 Paper 1 Offline
Numerical
+4
-0
An infinitely long uniform line charge distribution of charge per unit length $$\lambda$$ lies parallel to the y-axis in the y-z plane at $$z = {{\sqrt 3 } \over 2}$$a (see figure). If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the x-y plane with its centre at the origin is $${{\lambda L} \over {n{\varepsilon _0}}}$$ ($${{\varepsilon _0}}$$ = permittivity of free space), then the value of n is

JEE Advanced 2015 Paper 1 Offline Physics - Electrostatics Question 27 English
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4
IIT-JEE 2012 Paper 1 Offline
Numerical
+4
-0

An infinitely long solid cylinder of radius R has a uniform volume charge density $$\rho$$. It has a spherical cavity of radius R/2 with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is at a distance 2R from the axis of the cylinder, is given by the expression $${{23\rho R} \over {16k{\varepsilon _0}}}$$. The value of k is _____________.

IIT-JEE 2012 Paper 1 Offline Physics - Electrostatics Question 23 English

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