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A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is
In Millikan’s oil drop experiment an oil drop carrying a charge Q is held stationary by a p.d. 2400 v between the plates. To keep a drop of half the radius stationary the potential differ- ence had to be made 600 v. What is the charge on the second drop ?
An electric dipole is placed along the x-axis at the origin o. A point P is at a distance of 20 cm from this origin such that OP makes an angle $ \pi / 3$ with the x-axis. If the electric field at P makes an angle $ \theta $ with the x-axis, the value of $ \theta $ would be ...........
A particle having a charge of $ 1.6 \times 10^{-19} C $ enters between the plates of a parallel plate capaciter. The initial velocity of the particle is parallel to the plates. A potential difference of 300v is applied to the capacitor plates. If the length of the capacitor plates is 10cm and they are separated by 2cm, Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is $ 12 \times 10^{–24} kg $ .
If electron in ground state of H-atom is assumed in rest then dipole moment of electron proton system of H-atom is ............... Orbit radius of H atom in ground state is 0.56 $ A ^\circ $
$ P = e \times r0 $ calculate P
An electric dipole coincides on z axis and its mid point is on origin of the cartesian co-ordinate system. The electric field at an axial point at a distance z from origin is $ \bar E (z) $ and electric field at an equatorial point at a distance y from origin is $ \bar E (y) = | {\bar E(x) \over \bar E(y)}| ( y = z \lt \lt a ) = $
Theory related question
An oil drop of 12 excess electrons is held stationary under a constant electric field of $ 2.55 \times 10^4 Vm^{–1} $ . If the density of the oil is $ 1.26 gm/cm^3$ then the radius of the drop is
As the drop is stationary , weight of drop = force due to electric field $ { 4 \over 3 } \pi r^3 \rho g = neE $ So, $ r^3 = { 3neE \over 4 \pi \rho g } $ now find out r
Two points are at distances a and b (a < b) from a long string of charge per unit length $ \lambda $ The potential difference between the points in proportional to
$ E = - { dv \over dr } $ $ dv = { - \lambda \over 2 \pi \varepsilon_o r } $ $ { \lambda \over 2 \pi \varepsilon_o r } = { dv \over dr } $ $ \int _{va} ^ { vb } dv = -{ \lambda \over 2 \pi \varepsilon_0 } \int _a^b { 1 \over r } dr $
Two Points P and Q are maintained at the Potentials of 10 v and –4 v, respectively. The work done in moving 100 electrons from P to Q is
$ W = 100 e (-4 -10 ) = -1400 ev = -1400 (-1.6 \times 10^{-19} ) J = 2.24 \times 10^{-16} J $
Charges of $ + {10 \over 3 } \times 10^{-9} C $ are placed at each of the four corners of a square of side 8cm. The potential at the intersection of the diagonals is ....
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.