An oil drop of 12 excess electrons is held stationary under a constant electric field of 2 . 55 × 10 4 N / C (Millikan’s oil drop experiment). The density of the oil is 1 . 26 gm cm 3 . Estimate the radius of the drop. ( g = 9 . 8 m / s 2 ; e = 1 . 6 × 10 - 19 C ).

Given, Excess electrons on an oil drop, n = 12 Electric field intensity, E = 2.55 × 10 4 N C - 1 Density of oil, ρ = 1 .26 gm / cm 3 = 1 .26 × 10 3 kg/m 3 Acceleration due to gravity, g = 9.8 m s - 2 Charge on an electron, e = 1.6 × 10 - 19 C The total charge of the drop then becomes, q = n e = 12 × 1 . 6 × 10 - 19 C = 19 . 2 × 10 - 19 C Say, the radius of the sphere is r . The volume of the sphere then becomes, v = 4 3 π r 3 The mass of the sphere becomes, m = v ρ = 4 3 π r 3 ρ As the charge is held stationary under the electric field, the net force on a drop must be zero. Therefore, the force due to the electric field must be equal to the force due to gravity. ⇒ q E = m g ⇒ q E = 4 3 π r 3 ρ r = 3 E q 4 π ρ g 3 = 3 × 2.55 × 10 4 × 19 . 2 × 10 - 19 4 × 3.14 × 1.26 × 10 3 × 9.8 3 = 946.09 × 10 - 21 3 = 9.82 × 10 - 7 m

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

E M B I B E

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 28

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12th Meghalaya Board

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Derive an expression for work done in rotating an electric dipole in a uniform electric field.

Important Questions on Electric Charges and Fields

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12th Meghalaya Board

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

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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