Electric Charges and Fields

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Question
CBSEENPH12039977

A beam of cathode rays is subjected to crossed Electric (E)  and magnetic field (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by  

(Where V is the potential difference between cathode and anode)

  • B2/2VE2

  • 2VB2/E2

  • 2VE2/B2

  • 2E2/2VB2

Solution

D.

2E2/2VB2

As the electron beam is not deflected,
then
Fm = Fe
Or BeV = Ee
Or v= E/B   .... (i) 
As the electron moves from the cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If V is the potential difference between the anode and cathode, then potential energy of the electron at cathode = eV. Also, the kinetic energy of the electron at anode = mv2 /2. According to law of conservation of energy
1 half space mv squared space equals space eV
or space
straight v equals space square root of fraction numerator 2 eV over denominator straight m end fraction end root space space space... space left parenthesis ii right parenthesis

From space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space have
square root of fraction numerator 2 eV over denominator straight m end fraction end root space equals space straight E over straight B
Or space straight e over straight m space equals space fraction numerator straight E squared over denominator 2 VB squared end fraction

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Question
CBSEENPH12039922

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will 

  • be reduced to half

  • remain the same

  • be doubled

  • increase four times

Solution

B.

remain the same

Total space flux space space equals space fraction numerator Net space enclosed space charge over denominator straight epsilon subscript straight o end fraction

Question
CBSEENPH12039858

A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are:

  • zero and fraction numerator straight Q over denominator 4 πε subscript straight o straight R squared end fraction

  • fraction numerator straight Q over denominator 4 πε subscript straight o straight R end fraction space a n d space z e r o
  • fraction numerator straight Q over denominator 4 πε subscript straight o straight R end fraction space a n d space fraction numerator Q over denominator 4 pi epsilon subscript o R squared end fraction
  • Both are zero

Solution

B.

fraction numerator straight Q over denominator 4 πε subscript straight o straight R end fraction space a n d space z e r o
Electric field inside, Einside = 0

Potential, Vinside = Vsurfacefraction numerator 1 over denominator 4 πε subscript straight o end fraction Q over R

Question
CBSEENPH11020950

A hollow cylinder has to charge q coloumb within it. Φ is the electric flux in a unit of voltmeter associated with the curved surface B, the flux linked with the plane surface A in a unit of voltmeter will be:

  • 1 half open parentheses q over epsilon subscript o minus capital phi close parentheses
  • fraction numerator straight q over denominator 2 straight epsilon subscript straight o end fraction
  • straight ϕ over 3
  • straight q over straight epsilon subscript straight o minus straight ϕ

Solution

A.

1 half open parentheses q over epsilon subscript o minus capital phi close parentheses

Apply Gauss's law to calculate the charge associated with plane surface A.
Gauss's law states that the the net electric flux through any clsoed surface is equal to the net charge inside the surface divided by εo.That is 
straight ϕ subscript total space equals space straight q over straight epsilon subscript straight o
Let space electric space flux space linked space with space surfaces space straight A comma space straight B space and space space straight C space are space straight ϕ subscript straight A comma space straight ϕ subscript straight B space and space straight ϕ subscript straight c space respectively. space
That space is space
straight ϕ subscript total space equals space straight ϕ subscript straight A space plus straight ϕ subscript straight B space plus straight ϕ subscript straight C
since space straight ϕ subscript straight C space equals space straight ϕ subscript straight A
therefore
2 straight ϕ subscript straight A space plus space straight ϕ subscript straight B space equals space straight ϕ subscript total space equals space straight q over straight epsilon subscript straight o
or
straight ϕ subscript straight A space equals space 1 half open parentheses straight q over straight epsilon minus straight ϕ subscript straight B close parentheses
But space straight ϕ subscript straight B space equals space straight ϕ space left parenthesis given right parenthesis
straight ϕ subscript straight A space equals space 1 half space open parentheses straight q over straight epsilon subscript straight o minus straight ϕ close parentheses

Question
CBSEENPH11020958

A square surface of side L m is in the plane of the paper. A uniform electric field straight E with rightwards arrow on top space left parenthesis straight V divided by straight m right parenthesis, also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is

  • EL squared divided by left parenthesis 2 straight epsilon subscript 0 right parenthesis
  • EL squared divided by 2
  • zero

  • EL squared

Solution

C.

zero

Electric flux left parenthesis straight ϕ subscript straight e right parenthesis is a measure of the number of field lines crossing a surface. The number of field lines passing through unit area (N/S) will be proportional to the electric field, or, 
                          straight N over straight S proportional to straight E space rightwards double arrow space space straight N proportional to space ES
The quantity ES is the electric flux through surface S.
As we have seen in the problem that, lines of force that enter the closed surface leave the surface immediately, so no electric flux is bound to the system. Hence, electric flux is zero.

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