An infinite line charge produces a field of 9 × 10 4 N C at a distance of 2 cm . Calculate the linear charge density. Take, ∈ 0 = 8 . 854 × 10 - 12 C 2 / N . m 2

Given, Distance, d = 2 c m = 0.02 m Electric Field, E = 9 × 10 4 N / C Say, the linear charge density is λ . The formula for electric field produced by infinite line charge is given as, E = λ 2 π ∈ 0 d ⇒ λ = 2 π ∈ 0 d E = 2 × 3 . 14 × 8 . 854 × 10 - 12 × 0 . 02 × 9 × 10 4 = 0 . 1 μC / m

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17 × 10 - 12 C / m 2 . What is E : a) in the outer region of the first plate b) in the outer region of the second plate and c) between the plates? Take, ∈ 0 = 8 .854 × 10 - 12 N - 1 C 2 m - 2 .

A and B are two parallel plates close to each other. Outer region of plate A is labeled as I , outer region of plate B is labeled as I I I , and the region between the plates, A and B , is labeled as I I . Charge density of plate A , σ = 17.0 × 10 - 12 C / m 2 In the regions, I and I I I , electric field E is zero because the fields due to the equal and opposite charges on the plates gets canceled. Electric field E in region I I is, E = σ ∈ 0 = 17 .0 × 10 - 12 8 .854 × 10 - 12 = 1.92 N / C . Therefore, electric field between the plates is 1.92 N / C .

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

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A point charge of $2.0 μC$ is at the center of a cubic Gaussian surface $9.0 cm$ on edge. What is the net electric flux through the surface? ( $\in_{0} = 8.854 \times 10^{- 12}$ )

Important Questions on Electric Charges and Fields

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PHYSICS PART-1 TEXTBOOK FOR CLASS XII>Chapter 1 - Electric Charges and Fields>EXERCISES>Q 1.2

A point charge causes an electric flux of

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

to pass through a spherical Gaussian surface of

10 cm

radius centered on the charge.
If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

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PHYSICS PART-1 TEXTBOOK FOR CLASS XII>Chapter 1 - Electric Charges and Fields>EXERCISES>Q 1.2

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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PHYSICS PART-1 TEXTBOOK FOR CLASS XII>Chapter 1 - Electric Charges and Fields>EXERCISES>Q 1.21

A conducting sphere of radius

10 cm

has an unknown charge. If the electric field

20 cm

from the center of the sphere is

1.5 \times 10^{3} \frac{N}{C}

and points radially inward, what is the net charge on the sphere?

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PHYSICS PART-1 TEXTBOOK FOR CLASS XII>Chapter 1 - Electric Charges and Fields>EXERCISES>Q 1.22

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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PHYSICS PART-1 TEXTBOOK FOR CLASS XII>Chapter 1 - Electric Charges and Fields>EXERCISES>Q 1.23

An infinite line charge produces a field of

9 \times 10^{4} \frac{N}{C}

at a distance of

2 cm

. Calculate the linear charge density. Take,

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

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PHYSICS PART-1 TEXTBOOK FOR CLASS XII>Chapter 1 - Electric Charges and Fields>EXERCISES>Q 1.24

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17 \times 10^{- 12} C / m^{2}$ . What is $E$ : a) in the outer region of the first plate b) in the outer region of the second plate and c) between the plates? Take, $\in_{0} = 8 .854 \times 10^{- 12} N^{- 1} C^{2} m^{- 2}$ .

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PHYSICS PART-1 TEXTBOOK FOR CLASS XII>Chapter 1 - Electric Charges and Fields>ADDITIONAL EXERCISES>Q 1.25

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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PHYSICS PART-1 TEXTBOOK FOR CLASS XII>Chapter 1 - Electric Charges and Fields>ADDITIONAL EXERCISES>Q 1.26

Which among the curves shown in the figure cannot possibly represent electrostatic field lines?

Question Image

A point charge of 2.0 μC is at the center of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface? (∈0 = 8.854×10-12)

Important Questions on Electric Charges and Fields

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A point charge of $2.0 μC$ is at the center of a cubic Gaussian surface $9.0 cm$ on edge. What is the net electric flux through the surface? ( $\in_{0} = 8.854 \times 10^{- 12}$ )