The identical point charges, q each, are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q .

= Let the two charges of + q each placed at point A and B at a distance 2 m apart in air. Suppose, the third charge Q (unknown magnitude and charge) is placed at a point O, on the line joining the other two charges, such that O A = x , O B = 2 - x . For the system to be in equilibrium, the net force on each charge must be zero. If we assume that charge Q placed at O is positive, the force on it may be zero. But the force on charge q at point A or B will not be zero. It is because , the forces on a charge q due to the other two charges (Q and q) will act in the same direction. If charge Q is negative, then the forces on vharge q due to the other two charges will act in opposite direction. Hence, Q will be negative in nature. For the charge ( - Q ) to be in equilibrium Force on charge ( - Q ) due to charge ( + q ) at point A should be equal and opposite to charge ( + q ) at B 1 4 πε 0 Q q x 2 = 1 4 πε 0 Q q 2 - x 2 On solving this we get : x = 1 m

An electric field is uniform and acts along + x direction in the region of positive x . It is also uniform with the same magnitude but acts in - x direction in the region of negative x . The value of the field is E = 200 N / C for x > 0 and E = - 200 N / C for x < 0 . A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x - axis so that one flat face is at x = + 10 cm and the other is at x = - 10 cm . Find the net outward flux through the cylinder.

At the surface 1 , the area vector of the surface and electric field vector both are in the negative x-direction. Hence, the electric flux through the surface 1 : ϕ 1 = E → . A → = 200 × 3 . 14 × 25 × 10 - 4 = 1 . 57 N m 2 C - 1 At the surface 2 , the area vector of the surface and electric field vector both are in the positive x-direction. The electric flux through the surface 2 : ϕ 2 = E → . A → = 200 × 3 . 14 × 25 × 10 - 4 = 1 . 57 N m 2 C - 1 The outward flux through surface 3 (curved surface of the cylinder) will be zero as the electric field and the area vector are perpendicular. The net outward flux through the cylinder: ϕ = 2 × 1 . 57 = 3 . 14 N m 2 C - 1 The net outward flux through the cylinder is 3 . 14 N m 2 C - 1 .

EASY

12th Karnataka Board

IMPORTANT

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Explain the term linear charge density. Write its SI unit.

Important Questions on Electric Charges and Fields

EASY

12th Karnataka Board

IMPORTANT

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

Explain why two lines of force cannot intersect at a point.

HARD

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 15

Three equal charges of

10 \times 10^{- 8} C

respectively, each located at the corners of a right triangle whose sides are

15 cm

20 cm

, and

25 cm

respectively. Find the force exerted on the charge located at the

90 °

angle.

HARD

12th Karnataka Board

IMPORTANT

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

Obtain an expression for intensity of electric field in end on position, i.e., axial position of an electric dipole.

HARD

12th Karnataka Board

IMPORTANT

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

Total charge $- Q$ is uniformly spread along the length of a ring of radius $R$ . A small test charge $+ q$ of mass $m$ is kept at the center of the ring and is given a gentle push along the axis of the ring. Obtain time period for its simple harmonic motion.

HARD

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 18

Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where

E = 0

) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

MEDIUM

12th Karnataka Board

IMPORTANT

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

Define electric field intensity. Write its SI unit. Write the magnitude and direction of electric field intensity due to an electric dipole of length

2 a

at the midpoint of the line joining the two charges.

EASY

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Electric Charges and Fields>Exercise>Q 20

Explain law of conservation of charge.

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12th Karnataka Board

IMPORTANT

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

An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

Explain the term linear charge density. Write its SI unit.

Important Questions on Electric Charges and Fields

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