A point charge of 5 × 10 - 6 C experiences a force of 2 × 10 - 3 N when kept in a uniform electric field of intensity E . Find E .

Given that charge = q = 5 × 10 - 6 C and experienced force = F = 2 × 10 3 N According to relation between electrostatic force and electric field intensity F = qE ⇒ E = F q = 2 × 10 - 3 5 × 10 - 6 = 2 × 10 3 5 = 400 NC - 1

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 8

MEDIUM

12th Rajasthan Board

IMPORTANT

Earn 100

Electric field lines are pictorial representation of electric field around a charge, Using Gauss's Law in electrostatics, derive an expression for the electric field due to an uniformly charges thin spherical shell at a point inside the shell.

Important Questions on Electric Charges and Fields

EASY

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 9

Write the expression for Electric intensity due to infinite long charged wire at a radial distance

r

from the wire.

EASY

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IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 14

Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure -

Question Image

EASY

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IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 16

What is the effect of speed of a charged particle on its charge and mass.

EASY

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 17

Discuss the methods of charging by induction.

HARD

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 19

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?

MEDIUM

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 21

An electric dipole having dipole moment $4 \times 10^{- 9} C m$ is oriented at $30^{°}$ from the direction of a uniform electric field of magnitude $5 \times 10^{4} {NC}^{- 1}$ . Calculate the magnitude of the torque on the dipole.

MEDIUM

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 22

A point charge of

5 \times 10^{- 6} C

experiences a force of

2 \times 10^{- 3} N

when kept in a uniform electric field of intensity

E

. Find

E

.

HARD

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 24

Explain the concept of electric field. Express electric flux through a surface in terms of electric field intensity. Show that the electric flux through a cylindrical surface with its axis parallel to a uniform electric field is zero.

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