Electric Forces & Fields Test

Result

Electric Forces & Fields Test - 8

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Weekly Quiz Competition

Question 1
4 / -1

Two thin wires rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and –q. The potential difference between the centres of the two rings is

A

B

C
zero

D
1

Solution
Question 2
4 / -1

A charged particle q is shot towards another charged particle Q which is fixed, with a speed v it approaches Q upto a closest distance r and then returns. If q were given a speed 2v, the closest distances of approach would be

A
r

B
2r

C
r/2

D
r/4

Solution

By principle of conservation of energy
Question 3
4 / -1

From a uniform circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:

A
37/9 MR²

B
4MR²

C
40/9 MR²

D
10 MR²

Solution

let σ be the mass per unit area.

The total mass of the disc = σ x πR² = 9M
The mass of the circular disc out

Let us consider the above system as a complete disc of mass 9M and a negative mass M superimposed on it. Moment of Inertia (I₁₎ of the complete disc (9MR²)/2 about an axis passing through O and perpendicular to the plane of the disc.

M.I. of the cut-out portion about an axis passing through O and perpendicular to the plan of disc

Therefore, M.I (I2) of the cut out portion about an axis passing through O and perpendicular to the plane of disc

Using perpendicular axis theorem

Therefore, the total M.I. of the system about an axis passing through O and perpendicular to the plane of the disc is
I = I₁ + I₂
Question 4
4 / -1

The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field will be given by

A

B

C

D
Zero

Solution
Question 5
4 / -1

In a coil of resistance 10Ω, the induced current developed by changing magnetic flux through it is shown in the figure as a function of time. The magnitude of change in flux through the coil in weber is

A
8

B
2

C
6

D
5

Solution
Question 6
4 / -1

A charge q is placed at the corner of a cube of side a. The electric flux passing through the cube is :

A

B

C

D

Solution

According to Gauss's law, the electric flux through a closed surface is equal to 1/ε₀ times the net charge enclosed by the surface.

Since, q is the charge enclosed by the surface, then the electric flux ϕ = q/ε₀.

If charge q is placed at a corner of cube, it will be divided ito 8 such cubes. Therefore, electric flux through the cube is

Therefore, electric flux through the cube is ϕ′= 1/8 (q/ε₀)
Question 7
4 / -1

Two thin dielectric slabs of dielectric constants K₁ and K₂ (K₁ < K₂) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field E between the plates with distance d as measured from plate P is correctly shown by:

A

B

C

D
None

Solution

The electric field inside the dielectrics will be less than the electric field outside the dielectrics. The electric field inside the dielectric could not be zero.

As K₂ > K₁ the drop in electric field for K₂ dielectric must be more than K₁.
Question 8
4 / -1

A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are :

A

B

C

D
Both are zero

Solution

For a conducting sphere

Electric fidld at centre = 0
Question 9
4 / -1

In a region, the potential is represented by V(x,y,z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, z are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1,1) is :

A
6√5 N

B
30 N

C
24 N

D

Solution
Question 10
4 / -1

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will :

A
be doubled

B
increase four times

C
be reduced to half

D
remain the same

Solution

Method 1: Using Number of Field Lines

∙ We know that, Electric flux is proportional to number of field lines.

∙  Form figure, we see that the number of field lines crossing both the spherical surfaces is same.

∙  Therefore, the flux passing both the surfaces will be same.

Method 2: Using Gauss Law

∙ Gauss Law states that, Electric flux Δϕ = q_in/ϵ₀

∙ Assuming both the spherical surfaces S₁ and S₂ as gaussian surfaces. From figure, we find that the charge enclosed(q_in=Q), same for both the surfaces irrespective of radius.

Note:  We can observe that whatever be the size, shape and position of second surface, If it is enclosing the same charge, then the Number of field lines crossing it or charge enclosed by it remains the same as first surface.

Hence answer will be same, even if the second surface is cube or cylinder of any size.