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

Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.

Consider an electric dipole consisting of charges - q and + q which are separated by a distance of 2 a with a point O being the centre. Let a point P be at the axial line of dipole whose electric field is to be calculated. Let r be the distance between P and O . Let E A be the electric field due to charge - q and E B be the electric field due to charge + q . E A = 1 4 π ε 0 q A P 2 ⇒ 1 4 π ε 0 q a + r 2 E B = 1 4 π ε 0 q B P 2 ⇒ 1 4 π ε 0 q r - a 2 Since E A and E B are opposite to each other and E B > E A , then the resultant intensity at P would be, E = E B - E A ⇒ 1 4 π ε 0 q r - a 2 - 1 4 π ε 0 q r + a 2 ⇒ q 4 π ε 0 1 r - a 2 - 1 r + a 2 ⇒ q 4 π ε 0 r + a 2 - r - a 2 r 2 - a 2 2 ⇒ q 4 π ε 0 4 r a r 2 - a 2 2 ⇒ q 4 π ε 0 2 a 2 r r 2 - a 2 2 ⇒ 1 4 π ε 0 2 p r r 2 - a 2 2 Also, if r > > > a , then, E = 1 4 π ε 0 2 p r 3 .

State Gauss Law of electrostatics and explain it's importance.

Gauss's law states that total electric flux through any closed surface is equal to 1 ∈ 0 times the net charge enclosed by the surface. Total flux ϕ = ∮ E . → d s → = q ∈ 0 Importance : (1) Gauss's law is very useful in calculating the electric field in case of problems where it is possible to construct a closed surface by making a Gaussian surface. (2) Gauss's law is true for any closed surface, no matter what its shape or size. (3) Symmetric considerations in many problems make the application of Gauss's law much easier.

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IMPORTANT

Earn 100

Derive the equation for the couple acting on a electric dipole in a uniform electric field.

Important Questions on Electric Charges and Fields

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IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4

Consider a uniform electric field

E = 3 \times 10^{3} N / C

. What is the flux of this field through a square of

10 cm

on a side whose plane is parallel to the

y z

plane?

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IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4.2

The electrostatic force on a small sphere of charge

0.4 µ C

due to another small sphere of charge

- 0.8 µ C

in the air is

0.2 N .

What is the force on the second sphere due to the first?

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IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4.4

Explain the meaning of the statement ‘electric charge of a body is quantized’.

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Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4.12

Two insulated charged copper spheres

A

and

B

have their centers separated by a distance of

50 cm

. What is the mutual force of electrostatic repulsion if the charge on each is

6.5 \times 10^{- 7} C

?
The radii of

A

and

B

are negligible compared to the distance of separation. (Take,

\frac{1}{4 {πε}_{0}} = 9 \times 10^{9}

)

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IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4.2

A point charge causes an electric flux of

- 1.0 \times 10^{3} {Nm}^{2} C^{- 1}

to pass through a spherical Gaussian surface of

10 cm

radius centered on the charge. What is the value of the point charge? (

\in_{0} = 8.854 \times 10^{- 12} N^{- 1} m^{- 2} C^{2}

)

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IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4.22

A uniformly charged conducting sphere of

2.4 m

diameter has a surface charge density of

80 μC / m^{2}

. Find the charge on the sphere. What is the total electric flux leaving the surface of the sphere?

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12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4.25

An oil drop of

12

excess electrons is held stationary under a constant electric field of

2.55 \times 10^{4} N / C

(Millikan’s oil drop experiment). The density of the oil is

1.26 \frac{gm}{{cm}^{3}}

. Estimate the radius of the drop. (

g = 9.8 m / s^{2}

;

e = 1.6 \times 10^{- 19} C

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IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4.31

It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by

u

) of charge

+ (2 / 3) e,

and the ‘down’ quark (denoted by

d

) of charge

(- 1 / 3) e,

together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Derive the equation for the couple acting on a electric dipole in a uniform electric field.

Important Questions on Electric Charges and Fields

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