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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 is placed at an angle of $60^ \circ $ with an electric field of intensity $ 10^5 NC^{–1} $ . It experiences a torque equal to 83Nm . If the dipole length is 2cm then the charge on the dipole is c.
$ \tau = PE sin \theta = q (2q) E sin \theta $ $ q = { \tau \over (2a ) E sin \theta } $ Now find out q
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
A Charge q is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is
Theory related question
An infinitly long thin straight wire has uniform linear charge density of 1/3 c/m . Then, the magnitude of the electric intiensity at a point 18 cm away is ……$NC^–1$.
$ \int \bar E . \bar da = { \Sigma q \over \varepsilon o } $ $ E . 2 \pi r l = { q \over \varepsilon _0 } $ $ E = { q \over 2 \pi \varepsilon_o r l } $ $ E = { 2 \lambda \over 4 \pi \varepsilon_0 r } $ Now find out E
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 $
A long string with a charge of per unit length passes through an imaginary cube of edge l. The maximum possible flux of the electric field through the cube will be
Maximum length of string =$ \sqrt 3 l $ Maximum enclosed charge = $ \sqrt l \lambda $ $ \therefore \phi = { \sqrt 3 l \lambda \over \epsilon _o } $
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.