Test: Electric Charges & Fields - JEE 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.

As aluminium and copper are metals, their mobile electrons move under the influence of the external field until they reach the surface of the metal and collect there. External electric fields induce surface changes on metal objects that exactly cancel the field within. Since the field applied is same in both case, the induced charge will be the same.
Hence, the magnitude of induced charge in the aluminium will be the equal to that of copper.

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

Since the charge –q is moving in elliptical orbit so to make its motion stable the total angular momentum of the charge is constant since it experience a centripetal force from the charge +Q so it follow the motion as the motion of earth around sun.

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,

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 

The electrostatic force is an attractive and repulsive force between particles are caused due to their electric charges. The electric force between stationary charged bodies is conventionally known as the electrostatic force. 

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

In a uniformly charged hollow conducting sphere

Conductors are materials that permit electrons to flow freely from particle to particle. An object made of a conducting material will permit a charge to be transferred across the entire surface of the object.

Electric field inside the charged spherical shell is zero as there is no charge inside it.