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In a hydrogen atom a proton is fixed and an electron revolves
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In a hydrogen atom a proton is fixed and an electron revolves around it in an orbit of radius . The electrostatic attraction provides the centripetal force. Find speed of the electron. Mass of an electron is and charge on it is .

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Book:

Physics for JEE Main and Advanced Electrostatics and Current Electricity (McGraw Hill)

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In a hydrogen atom a proton is fixed and an electron revolves around it in an orbit of radius . The electrostatic attraction provides the centripetal force. Find speed of the electron. Mass of an electron is and charge on it is .
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Practice questions from Physics for JEE Main and Advanced Electrostatics and Current Electricity (McGraw Hill)

Question 1
Medium
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Views: 5,399

A hemisphere of radius is placed in a uniform electric field of strength . Find the electric flux through the curved surface if (a) The field is parallel to the base (b) Perpendicular to the base
Topic

Topic:

Electric Charges and Fields

Book

Book:

Physics for JEE Main and Advanced Electrostatics and Current Electricity (McGraw Hill)

View 4 solutions

Question 2
Medium
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Views: 6,074

Two identical positive charges are kept at a separation . is a point on the perpendicular bisector of the line joining the two charges. Distance is . Find electric field at point . Approximate your answer for .
Topic

Topic:

Electric Charges and Fields

Book

Book:

Physics for JEE Main and Advanced Electrostatics and Current Electricity (McGraw Hill)

View solution

Question 3
Medium
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Views: 5,837

Particle having positive charge is moving directly head-on towards initially stationary positively charged particle . At the instant when and are closest together
  1. the momenta of and must be equal
  2. the velocities of and must be equal
  3. would have gained less kinetic energy than would have lost.
  4. would have gained the same momentum as would have lost
Topic

Topic:

Electric Charges and Fields

Book

Book:

Physics for JEE Main and Advanced Electrostatics and Current Electricity (McGraw Hill)

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Question 4
Medium
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Views: 5,725

A small ball has a mass and carries a charge . It is thrown from a horizontal surface at a speed of making an angle of with the horizontal. The entire space has a horizontal electric field N/C. At what distance from the point of projection will the particle again hit the horizontal surface.
Topic

Topic:

Electric Charges and Fields

Book

Book:

Physics for JEE Main and Advanced Electrostatics and Current Electricity (McGraw Hill)

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Practice more questions from Electric Charges and Fields

Question 1
Easy
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Views: 7,812

Two large charged plane sheets of charge densities and are arranged vertically with a separation of between them. Deduce expressions for the electric field at points (i) to the left of the first sheet, (ii) to the right of the second sheet, and (iii) between the two sheets.
Topic

Topic:

Electric Charges and Fields

Exam

Exam:

Class 12th Board 2019

View 22 solutions

Question 2
Medium
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Views: 5,578

A conducting sphere of radius having charge is placed in a uniform external field . is the centre of the sphere and is a point on the surface of the sphere such that makes an angle of with the opposite direction of external field. Calculate the potential at point due to charge on the sphere only.
Topic

Topic:

Electric Charges and Fields

Book

Book:

Problems in Physics II for IIT JEE (McGraw Hill)

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Question 3
Medium
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Views: 5,397

Four charges are arranged at the corners of a square as shown in the figure. The force on the charge kept at the centre is
  1. zero
  2. along the diagonal
  3. along the diagonal
  4. along perpendicular to the side
Topic

Topic:

Electric Charges and Fields

Book

Book:

Master Resource Book in JEE Main Physics (Arihant)

View 25 solutions

Question 4
Medium
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Views: 5,771

An electron of mass , initially at rest, moves through a certain distance in a uniform electric field in time . A proton of mass also initially at rest, takes time to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio is nearly equal to
  1. 1
  4. 1836
Topic

Topic:

Electric Charges and Fields

Book

Book:

Master Resource Book in JEE Main Physics (Arihant)

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Practice questions on similar concepts asked by Filo students

Question 1
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Title: Analysis of Signals and Systems in MATLAB Consider the following three systems: System 1: w[n] = x[n] - x[n-1] - x[n-2] y[n] = cos(x[n]) z[n] = n * x[n] System 2: w[n] = x[n] y[n] = cos(x[n]) z[n] = n * x[n] System 3: w[n] = x[n] y[n] = cos(x[n]) z[n] = n * x[n] where x[n] is the input to each system, and w[n], y[n], and z[n] are the corresponding outputs. (a). Consider the three input signals x1[n] = 8[n], x2[n] = 8[n-1], and x3[n] = (8[n] + 2 * [n-1]). For System 1, store in w1, w2, and w3 the responses to the three inputs. The vectors w1, w2, and w3 need to contain the values of w[n] only on the interval 0 < n < 5. Use subplot and stem to plot the four functions represented by w1, w2, w3, and w1 + 2 * w2 within a single figure. Make analogous plots for Systems 2 and 3. (b). State whether or not each system is linear. If it is linear, justify your answer. If it is not linear, use the signals plotted in Part (a) to supply a counter-example. (c). State whether or not each system is time-invariant. If it is time-invariant, justify your answer. If it is not time-invariant, use the signals plotted in Part (a) to supply a counter-example.
Topic

Topic:

Physics

View solution

Question 2
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Views: 5,662

density of iron is .
Topic

Topic:

Mechanics

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Question 3
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Views: 6,008

The potential due to a uniformly charged rod of length / and charge at point at a distance of from one end as shown in the diagram is Clear Response
Topic

Topic:

Electrostats

View solution

Question 4
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Views: 5,148

5 Potential due to a Group of Point Charges Potential is a scalar quantity. Therefore, the potential at any point due to a group of point-charges is found by calculating the potential due to each charge (as if the other charges were not present), and then adding algebraically the quantities so obtained. Thus, if a point is at distances and metre from the point charges and coulomb respectively, then the resultant potential at that point will be \[ V=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r_{2}}-\frac{q_{3}}{r_{3}}-\frac{q_{4}}{r_{4}}\right] \text { volt. } \] If there are point charges, the potential due to them at a point will be where is the distanco of the point from the charge . If the charge distributidn be continuous, then the summation in the above expression will be replaced by integration, that is, where is a differentia element of the charge distribution and is its distance from the point at which is to be calculated. If the charge is spread cottinuously over an area , then , where is surface density of charge. Then, we have where is a surface integfal. Similarly, if the charge is dptributed continuously within a volume , then where is volume density of charge and is a volume integral.
Topic

Topic:

Electrostats

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Question Text
In a hydrogen atom a proton is fixed and an electron revolves around it in an orbit of radius . The electrostatic attraction provides the centripetal force. Find speed of the electron. Mass of an electron is and charge on it is .
TopicElectric Charges and Fields
SubjectPhysics
ClassClass 12
Answer TypeText solution:1
Upvotes68