Electric Charges and Fields Test

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Electric Charges and Fields Test - 7

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Weekly Quiz Competition

Question 1
4 / -1

The SI unit of electric flux is

A
N C^-1 m²

B
N C m^-2

C
N C^-2 m²

D
N C^-1 m^-2

Solution

As electric flux, φ = E.Δs
∴ unit of φ is N C^-1 m²
Question 2
4 / -1

The dimensional formula of electric flux is

A
[M¹L¹T^-2]

B
[M¹L³T^-3A^-1]

C
[M²L²T^-2A^-2]

D
[M¹L^-³T³A¹]

Solution

Electric flux, φ=
The dimension of φ = Dimension of E x dimension of s
[φ] = [M¹L¹T^-2][AT]^-1[L²] = [M¹L³T^-3A^-1]
Question 3
4 / -1

A circular plane sheet of radius 10 cm is placed in a uniform electric field of 5 x 10⁵N C^-1, making an angle of 60⁰ with the field. The electric flux through the sheet is

A
1.36 x 10² N m² C^-1

B
1.36 x 10⁴ N m² C^-1

C
0.515 x 10² N m² C^-1

D
0.515 x 10⁴N m² C^-1

Solution

Here, r = 10 cm 0.1 m, E = 5 x 10⁵ N C^-1
As the angle between the plane sheet and the electric field is 60⁰, angle made by the normal to the plane sheet and the electric field is θ = 90⁰ - 60⁰ = 30⁰
φ_E = ES cos θ = E x πr² cos θ
= 5 x 10⁵ x 3.14 x (0.1)² cos 30⁰ = 1.36 x 10⁴ N m²C^-1
Question 4
4 / -1

A uniform electric field E = 2 x 10³ N C^-1 is acting along the positive x-axis. The flux of this field through a square of 10cm on a side whose plane is paralled to the yz plane is

A
20 N C^-1 m²

B
30 N C^-1 m²

C
10 N C^-1 m²

D
40 N C^-1 m²

Solution

Here, E = 2 x 10³ N C^-1 is along + x-axis
Surface area, s = (10 cm)² = 10²x 10^-4 m² = 10^-2 m²
When plane is parallel to yz plane, θ = 0°
So φ = E s cosθ = 2 x 10³ x 10^-2 cos 0^° = 20N C^-1 m²
Question 5
4 / -1

In the question number 66, the flux through the same square if the line normal to its plane makes a 60° angle with the x-axis is

A
30 N C^-1 m²

B
10 N C^-1 m²

C
20 N C^-1m²

D
25 N C^-1 m²

Solution

When normal to the plane makes an angle of 60° with x-axis, then θ = 60°
φ = Es cosθ = 2 x 10³ x 10^-2 cos 60° = 10 N C^-1N C^-1m²
Question 6
4 / -1

Which of the following statements about dipole moment is not true?

A
The dimensions of dipole moment is [LTA].

B
The unit of dipole moment of C m.

C
Dipole moment is vector quantity and directed from negative to positive charge.

D
Dipole moment is a scalar quantity and has magnitude charge equal to the potential of separation between charge.

Solution

Dipole moment is a vector quantity and has magnitude of 2qa and it is in the direction of the dipole axis from -q to q.
Question 7
4 / -1

The unit of electric dipole moment is

A
newton

B
coulomb

C
farad

D
debye

Solution

The unit of electric dipole moment is: 2ql = C − m = Debye
Question 8
4 / -1

Two charges + 20 μC and -20 μC are placed 10 mm apart. The electric field at point P, on the axis of the dipole 10 cm away from its centre O on the side of the positive charge is

A
8.6 x 10⁹ N C^-1

B
4.1 x 10⁶ N C^-1

C
3.6 x 10⁶N C^-1

D
4.6 x 10⁵ N C^-1

Solution

Here, q = ±20μC = ±20 x 10^-6C, 2a = 10 mm = 10 x 10^-3m
r = OP = 10 cm = 10 x 10^-2 m
= q x 2a = 20 x 10^-6 x 10 x 10^-3 m = 2 x 10^-7 m
The electric field along BP,
As a < < r,
Question 9
4 / -1

Consider a region inside which, there are various types of charges but the total charge is zero. At points outside the region

A
the electric field is necessarily zero.

B
the electric field is due to the dipole moment of the charge distribution only.

C
the dominant electric field is inversely proportional to r³, for large r (distance from origin).

D
the work done to move a charged particle along a closed path, away from the region will not be zero.

Solution

When there are various types of charges in a region, but the total charge is zero, the region can be supposed to contain a number of electric dipoles. Therefore, at points outside the region, the dominant electric field ∝ (1/r³) for large r.
Question 10
4 / -1

Two point charges of 1μC and -1μC are separated by a distance of 100 Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be

A
9 N C^-1

B
0.9 N C^-1

C
90 N C^-1

D
0.09 N C^-1

Solution

The Point lies on equatorial line of a short dipole,

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Re-Attempt Weekly Quiz Competition

Electric Charges and Fields Test - 7

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Electric Charges and Fields Test - 7

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Question 1

N C-1 m2

N C m-2

N C-2 m2

N C-1 m-2

Solution

Question 2

[M1L1T-2]

[M1L3T-3A-1]

[M2L2T-2A-2]

[M1L-3T3A1]

Solution

Question 3

1.36 x 102 N m2 C-1

1.36 x 104 N m2 C-1

0.515 x 102 N m2 C-1

0.515 x 104 N m2 C-1

Solution

Question 4

20 N C-1 m2

30 N C-1 m2

10 N C-1 m2

40 N C-1 m2

Solution

Question 5

30 N C-1 m2

10 N C-1 m2

20 N C-1 m2

25 N C-1 m2

Solution

Question 6

The dimensions of dipole moment is [LTA].

The unit of dipole moment of C m.

Dipole moment is vector quantity and directed from negative to positive charge.

Dipole moment is a scalar quantity and has magnitude charge equal to the potential of separation between charge.

Solution

Question 7

newton

coulomb

farad

debye

Solution

Question 8

8.6 x 109 N C-1

4.1 x 106 N C-1

3.6 x 106 N C-1

4.6 x 105 N C-1

Solution

Question 9

the electric field is necessarily zero.

the electric field is due to the dipole moment of the charge distribution only.

the dominant electric field is inversely proportional to r3, for large r (distance from origin).

the work done to move a charged particle along a closed path, away from the region will not be zero.

Solution

Question 10

9 N C-1

0.9 N C-1

90 N C-1

0.09 N C-1

Solution

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N C^-1 m²

N C m^-2

N C^-2 m²

N C^-1 m^-2

[M¹L¹T^-2]

[M¹L³T^-3A^-1]

[M²L²T^-2A^-2]

[M¹L^-³T³A¹]

1.36 x 10² N m² C^-1

1.36 x 10⁴ N m² C^-1

0.515 x 10² N m² C^-1

0.515 x 10⁴N m² C^-1

20 N C^-1 m²

30 N C^-1 m²

10 N C^-1 m²

40 N C^-1 m²

30 N C^-1 m²

10 N C^-1 m²

20 N C^-1m²

25 N C^-1 m²

8.6 x 10⁹ N C^-1

4.1 x 10⁶ N C^-1

3.6 x 10⁶N C^-1

4.6 x 10⁵ N C^-1

the dominant electric field is inversely proportional to r³, for large r (distance from origin).

9 N C^-1

0.9 N C^-1

90 N C^-1

0.09 N C^-1