Derive an expression for intensity of electric field at a point

Subject

Physics

Class

ICSE Class 12

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 Multiple Choice QuestionsShort Answer Type

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 Multiple Choice QuestionsLong Answer Type

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46.

Derive an expression for intensity of electric field at a point in broadside position or on an equatorial line of an electric dipole.


Electric field intensity at a point on the broad side-on position equatorial line:

Consider an electric dipole consisting of two equal but opposite charges -q and +q separated by a vector distance 2l.

Let P be a point at a distance r from the centre of the dipole O.

The electric intensity at P due to the dipole is the vector sum of the field due to the charge -q at A and +q at B.



The resultant intensity is the vector sum of EA and EB.

EA and EB can be resolved into two components.

The y-components of the field cancel each other because,

EA sin straight theta = EB sin straight theta, oppositely directed.
The x components add up to give the resultant field E.

Magnitude of E is,

open vertical bar straight E with rightwards harpoon with barb upwards on top close vertical bar space equals space E subscript A space cos space theta space plus space E B space cos space theta

E space equals space fraction numerator 1 over denominator 4 pi epsilon subscript o end fraction. fraction numerator q over denominator open parentheses r squared plus l squared close parentheses end fraction cos space theta space plus space
space space space space space space space space space fraction numerator 1 over denominator 4 pi epsilon subscript o end fraction. fraction numerator q over denominator open parentheses r squared plus l squared close parentheses end fraction cos space theta space
space space space equals space 2 space straight x space fraction numerator 1 over denominator 4 pi epsilon subscript o end fraction fraction numerator q over denominator open parentheses r squared plus l squared close parentheses end fraction cos space theta

For the right angled triangle OPB,

cos space theta space equals space fraction numerator O B over denominator P B end fraction equals fraction numerator 1 over denominator left parenthesis r squared plus l squared right parenthesis to the power of bevelled 1 half end exponent end fraction
The dipole moment,

p = 2lq

E = fraction numerator 1 over denominator 4 πε subscript straight o end fraction. space fraction numerator p over denominator left parenthesis r squared plus l squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction

E is directed along PC. The direction of E can be found out by drawing the line of force passing through the point P. 

The direction of e at P is opposite to the direction of the dipole moment p.

That is, parallel to the line joining the two charges and directed from +q to -q.

When r>>1, then l2 is very small as compared to r2.

Then,

straight E subscript eq space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction straight x straight p over open parentheses straight r squared close parentheses to the power of 3.2 end exponent

space space space space space space space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction straight x straight p over straight r cubed
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47.

Figure below shows a capacitor C, an inductor L and resistor R, connected in series to an ac supply of 220 V.


Calculate:

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