31 Years NEET Previous Year Questions: Electric Charges & Fields Free MCQ

Net force on each of the charge due to the other charges is zero. However, disturbance in any direction other than along the line on which the charges lie, will not make the charges return.

At equilibrium, net force is zero

Charges (q) = 2 × 10^-6 C,

Distance (d) = 3 cm = 3 × 10^-2 m

and electric field (E) = 2 × 10⁵ N/C. Torque (t) = q.d.

E = (2 × 10^-6) × (3 × 10^-2) × (2 × 10⁵)

= 12 × 10^-3N-m.

Charge (q) = 0.2 C; Distance (d) = 2m; Angle
θ = 60º and work done (W) = 4J.
Work done in moving the charge (W)
= F.d cos θ = qEd cos θ

or

Intensity of electric field due to a Dipole

Intensity of electric field due to a Dipole

Dipole is formed when two equal and unlike charges are placed at a short distance.

• When these charges are placed at a short distance apart, they create an electric field.
• This field influences other charges nearby, demonstrating the dipole's effects.
• If the distance is too long, the charges act more like individual charges rather than a dipole.
• Thus, a dipole is effectively formed by placing equal and unlike charges close together.

Given : Electric field (E) = 500 V/m
and potential difference (V) = 3000 V.
We know that electric field

[where d = Distance between point charge
and A]

Charge resides on the outer surface of a conducting hollow sphere of radius R. We consider a spherical surface of radius r < R.
By Gauss's theorem,

i.e. electric field inside a hollow sphere is zero.

K.E. = Force × distance = qE.y

in air 

Let the semi-circular arc of radius aaa be in the xy-plane, centre at O, with the arc in the upper half so that it is symmetric about the y-axis.
Linear charge density on the arc is λ.

Take a small element of the arc at angle θ from the y-axis.
Length of the element: a dθ

Charge on element: ​

Distance of the element from O is a.
Electric field at O due to dq has magnitude:

Resolve dE into x- and y-components.

• Because of symmetry, the xxx-components from elements at +θ and −θ cancel.

• The y-components add.

For our choice of θ (from the y-axis),

Now,

For a semicircle, θ varies from 

So, 

Direction: along the negative y-axis (towards the straight diameter of the semicircle).

​​

The electric flux through the six faces of a cube with a charge Q at one corner can be determined using Gauss's Law. According to this law, the total electric flux (Φ) through a closed surface is proportional to the enclosed charge.

In this scenario, since the charge is located at a corner, only a fraction of the total flux is associated with the cube. Specifically:

• The total flux due to charge Q is given by: Φ = Q / ε₀.
• Since the cube represents only one-eighth of the total flux (due to symmetry), the flux through the cube is:
• Φ_cube = (1/8) × (Q / ε₀) = Q / (8ε₀).

Thus, the electric flux passing through all the six faces of the cube is:

Q / 8ε₀.

Given : Length of the dipole (2l) =10cm
= 0.1m or l = 0.05 m
Charge on the dipole (q) = 500 μC = 500 × 10^–6 C and distance of the point on the axis from the mid-point of the dipole (r) = 20 + 5 = 25 cm = 0.25 m. We know that the electric field intensity due to dipole on the given 

Given : Dipole moment of the dipole =  and
uniform electric field = . 
We know that dipole moment (p) = q.a (where q is the charge and a is dipole length). And when a dipole of dipole moment  is placed in a uniform electric field  , then Torque (τ) = Either force × Perpendicular distance between the two forces = qaEsinθ or τ = pE sin θ or 

Total flux out of all six faces as per Gauss's
theorem should be 

Therefore, flux coming out of each face

Charges (–e) on electron and (e) on proton exert a force of attraction given by

Note : Magnitude of Coulomb force is given by 

but in vector form 

Cube has 6 faces. Flux through any face is given by

Flux =

is electric field vector & is area vector

Here, angle between  is 90º.

Work done in rotating a dipole = pE (1 – cos θ)

If θ = 90º, work done = pE (1 – 0) = pE

According to Gauss Law state that the net electric through the closed surface {Gaussian's surface} is equal to net charges inside the surface divided by ∈₀
Here, Let flux on A , B and C surfaces are ФA, ФB, ФC
Then, Фnet = ФA + ФB + ФC = q/∈₀
We see that , ФA = ФC and ФB = Ф
So, 2ФA + Ф = q/∈₀
ФA = 1/2(q/∈₀ - Ф)
Hence, the electric flux through plane surface is 1/2(q/∈₀ - Ф)

Three point charges +q, –2q and +q are placed at points B (x = 0, y = a, z = 0), O (x = 0, y = 0, z = 0) and A(x = a, y = 0, z = 0)
The system consists of two dipole moment vectors due to (+q and –q) and again due to (+q and –q) charges having equal magnitudes qa units – one along 

 and other along   Hence, net dipole moment   along  at an angle 45° with positive X-axis. 

By the symmetry of the figure, the electric fields at O due to the portions AC and BD are equal in magnitude and opposite in direction. So, they cancel each other.

Similarly, the field at O due to CD and AKB are equal in magnitude but opposite in direction. Therefore, the electric field at the centre due to the charge on the part ACDB is E along OK.

Correct Answer is C.

Let n be the number of electrons missing.

Electric flux, φ = EA cos θ , where θ = angle between E and normal to the surface.

Electric field at a point inside a charged conducting spherical shell is zero.

By Gauss’s theorem,

Thus, the net flux depends only on the charge enclosed by the surface. Hence, there will be no effect on the net flux if the radius of the surface is doubled