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Electric Field: Electric Dipole

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Question
ICSEENIPH12029766
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(i) Write an expression (derivation not required) for intensity of electric field in:

(1) Axial position.

(2) Broad side position of an electric dipole, in terms of its length (2a) dipole moment (p) and distance (r).

Solution

1.
Intensity of electric field in axial position is: 
E1 = fraction numerator 1 over denominator 4 πε subscript straight o end fraction space open parentheses fraction numerator 2 P subscript r over denominator left parenthesis r squared space minus space a squared right parenthesis squared end fraction close parentheses ; p is the dipole moment
2.
Broad Side:
straight E subscript 2 space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space open square brackets fraction numerator straight P over denominator left parenthesis straight r squared space plus space straight a squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction close square brackets

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Question
ICSEENIPH12029767
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What is the ratio of axial electric field intensity and broad side electrcic field intensity E1: E2 for a short electric dipole?

Solution

For short dipole r > > a;
straight E subscript 1 space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space fraction numerator 2 straight P over denominator straight r cubed end fraction space straight E subscript 2 space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space straight P over straight r cubed Therefore comma space straight E subscript 1 colon space straight E subscript 2 space equals space 2 space colon 1

Question
ICSEENIPH12029799
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A point charge of 5 x 10-6 C experiences a force of 2 x 10-3 N when kept in a uniform electric field of intensity E. Find E.

Solution

Electric field Intensity,
E = fraction numerator 2 space straight x space 10 to the power of negative 3 end exponent over denominator 5 space straight x space 10 to the power of negative 6 end exponent end fraction space equals space fraction numerator 2 space x 10 cubed over denominator 5 end fraction equals 400 space V divided by m

Question
ICSEENIPH12029832
Wired Faculty App

ndeviated through a uniform electric field straight E with rightwards harpoon with barb upwards on top for intensity 5 x 104 Vm-1 and a uniform magnetic field straight B with rightwards harpoon with barb upwards on top.

(i) Find the magnitude of magnetic flux density B of the magnetic field.

(ii) What is the direction of straight B with rightwards harpoon with barb upwards on top space comma space if space space straight v with rightwards harpoon with barb upwards on top is towards right and straight E with rightwards harpoon with barb upwards on top is vertically downwards in the plane of this paper?

Solution

open vertical bar straight v with rightwards harpoon with barb upwards on top close vertical bar space equals space 2 space x space 10 to the power of 6 space m divided by s open vertical bar E close vertical bar space equals space 5 space x space 10 to the power of 4 space V space m to the power of negative 1 end exponent space minus space 1 space a n d space open vertical bar B with rightwards harpoon with barb upwards on top close vertical bar space equals space ? space straight i right parenthesis thin space Condition space under space which space the space lectron space pass space undeviated. straight tau space open vertical bar straight E with rightwards harpoon with barb upwards on top close vertical bar space equals space straight tau space open vertical bar straight v with rightwards harpoon with barb upwards on top close vertical bar open vertical bar straight B with rightwards harpoon with barb upwards on top close vertical bar space equals space 5 space straight x space 10 to the power of 4 space equals space 2 space straight x space 10 to the power of 6 space open vertical bar straight B with rightwards harpoon with barb upwards on top close vertical bar open vertical bar straight B with rightwards harpoon with barb upwards on top close vertical bar space equals space fraction numerator 5 over denominator 2 space straight x space 10 squared end fraction space equals space 2.5 space straight x space 10 to the power of negative 2 end exponent space wb divided by straight m squared
ii) Direction of straight B with rightwards harpoon with barb upwards on top is perpendicular to bold v with bold rightwards harpoon with barb upwards on top bold space a n d bold space bold E with bold rightwards harpoon with barb upwards on top using along the plane of the paper.

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