Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

Consider an electric dipole consisting of charges - q and + q separated by a distance 2 a . Let E A be the electric field at point Q due to charge - q and E B be the electric field at point Q due to charge + q where Q is a point on the equatorial line of the dipole whose electric field is to be measured. Let x be the distance between the centre of the dipole and point Q . Now, resultant electric field at point Q would be, E Q = E A + E B E A = 1 4 π ε 0 q A Q 2 ⇒ 1 4 π ε 0 q x 2 + a 2 ( ∵ According to Pythagoras, A Q = x 2 + a 2 2 ) E B = 1 4 π ε 0 q B Q 2 ⇒ 1 4 π ε 0 q x 2 + a 2 ( ∵ According to Pythagoras, B P = x 2 + a 2 2 ) Now, according to parallelogram law of vector addition, the resultant electric field would become, E Q = E A 2 + E B 2 + 2 E A E B cos 2 θ Now, since E A = E B we can say, E A = E B = E ' E Q = E ' 2 + E ' 2 + 2 E ' E ' cos 2 θ ⇒ 2 E ' 2 ( 1 + cos 2 θ ) ⇒ E ' 2 ( 1 + cos 2 θ ) ⇒ E ' 2 2 cos 2 θ E Q = 2 E ' cos θ - - - ( i ) Now, from △ A O Q , cos θ = a x 2 + a 2 Also, we know that, E ' = 1 4 π ε 0 q x 2 + a 2 Putting both these values in equation ( i ) , we get, E Q = 2 4 π ε 0 q x 2 + a 2 a x 2 + a 2 E Q = p 4 π ε 0 x 2 + a 2 3 2 Since we know that in dipoles, x > > > 2 a , then a 2 can be neglected. ∴ E = p 4 π ε 0 x 3 2

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 29

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On charging a soap bubble with negative charge its radius

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Important Questions on Electric Charges and Fields

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 30

How many electrons should be removed from a coin of mass

1.6 g

, so that it may float in an electric field of intensity

10^{9} N C^{- 1}

directed upward

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 31

Some point charges are placed on the circumference of circle at equal distance (see figure). The direction of electric field at centre $O$ will be along
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