1
JEE Advanced 2021 Paper 1 Online
Numerical
+2
-0
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Two point charges $$-$$Q and +Q/$$\sqrt 3 $$ are placed in the xy-plane at the origin (0, 0) and a point (2, 0), respectively, as shown in the figure. This results in an equipotential circle of radius R and potential V = 0 in the xy-plane with its center at (b, 0). All lengths are measured in meters.

JEE Advanced 2021 Paper 1 Online Physics - Electrostatics Question 26 English

The value of b is __________ meter.
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2
JEE Advanced 2020 Paper 2 Offline
Numerical
+3
-1
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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) JEE Advanced 2020 Paper 2 Offline Physics - Electrostatics Question 34 English
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3
JEE Advanced 2020 Paper 2 Offline
Numerical
+3
-1
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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) JEE Advanced 2020 Paper 2 Offline Physics - Electrostatics Question 35 English
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4
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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