Test: Electric Charges & Fields - NEET MCQ

Explanation:

When a negatively charged conductor is connected to the earth, electrons will flow from the conductor to the earth. This is because electrons have a negative charge and they will be repelled from the negatively charged conductor and attracted to the positively charged earth. As electrons flow from the conductor to the earth, the negative charge on the conductor will gradually decrease until it becomes neutral.

• Option A is incorrect because charge flow does occur when a negatively charged conductor is connected to the earth.
• Option B is incorrect because protons have a positive charge and they are not free to move in a conductor.
• Option C is incorrect because electrons flow from the earth to the conductor, not the other way around.

Electric flux, Φ = −1.0 × 10³ N m²/C

Radius of the Gaussian surface,

r = 10.0 cm

Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −10³ N m²/C.

Electric flux is given by the relation,

qϕ=q∈0

Where,

q = Net charge enclosed by the spherical surface

∈₀ = Permittivity of free space = 8.854 × 10⁻¹² N⁻¹ C² m⁻²

∴ qq=ϕ∈0

= −1.0 × 10³ × 8.854 × 10⁻¹²

= −8.854 × 10⁻⁹ C

= −8.854 nC

Therefore, the value of the point charge is −8.854 nC.

In case I :


In Case II :


From equations (i) and (ii),

Let m be mass of each ball and q be charge on each ball. Force of repulsion,


In equilibrium
Tcosq = mg ...(i)
Tsinq = F ...(ii)
Divide (ii) by (i), we get,

From figure (a),



Divide (iv) by (iii), we get

Acceleration 

This consists of two dipoles, –q and +q with dipole moment along with the +y-direction and –q and +q along the x-direction.


Along the direction 45° that is along OP, where P is (+a, +a, 0).

Here, q = 30°, E = 2 x 10⁵ N C^–1, τ = 4 N m,
l = 2 cm = 0.02 m, q = ?
τ = pE sinq = (ql)E sinq

Here, r = 10 cm, q = 3.2 x 10^–7 C

It is not possible to remove or add protons/neutrons to an atom, but electrons can be added or removed by an atom easily.
By adding electrons it becomes negatively charged.
By removing electrons it becomes positively charged. 

Let φ_A, φ_B and φ_C are the electric flux linked with surface is A, B and C.
According to Gauss theorem,

Let us consider a spherical shell of radius x and thickness dx.

Charge on this shell

∴ Total charge in the spherical region from centre to r(r<R) is

As at a corner, 8 cubes can be placed symmetrically, flux linked with each cube (due to a charge Q at the corner)
will be 

Now for the faces passing through the edge A, electric field E at a face will be parallel to area of face and so flux for these three faces will be zero. Now as the cube has six faces and flux linked with three faces (through A) is zero, so flux linked with remaining three faces will be 
Hence, electric flux passed through all the six faces of the cube is 

λ= linear charge density; Charge on elementary portion dx=λdx.

Electric field at 
Horizontal electric field, i.e., perpendicular to AO, will be cancelled. Hence, net electric field = addition of all electrical fields in direction of AO =ΣdEcosθ

Electric flux emerging from the cube does not depend on size of cube.

Electric Field Formula:
Electric field due to an infinitely long wire: 
Calculate Electric Fields:

Net Electric Field at Midpoint:
- Since the fields are in opposite directions, the net field is:
E = E₂ − E₁ = 7.2 × 10¹¹−3.6 × 10¹¹ = 3.6 × 10¹¹ N/C

In a uniformly charged hollow conducting sphere

Force acting on the charged particle due to electric


work done in moving through distance S,

Electric field intensity at the centre of the disc. E = σ/2ϵ₀ (given)
Electric field along the axis at any distance x from the centre of the disc

From question, x=R (radius of disc)

∴% reduction in the value of electric field