Test: MCQs (One or More Correct Option): Electrostatics | JEE Advanced - JEE MCQ

Let us consider the positive charge Q at any instant of time t at a distance x from the origin. It is under the influence of two forces   On resolving these two forces we find that F sin θ cancels out. The resultant force is

F_R = 2F cos θ

Since F_R is not proportional to x, the motion is NOT simple harmonic. The charge Q will accelerate till the origin and gain velocity. At the origin the net force is zero but due to momentum it will cross the origin and more towards left. As it comes on negative x-axis, the force is again towards the origin.   

q has to be negative for equilibrium.

Considering equilibrium of 1

Charge on plate is q

Charge on plate is q

The potential inside the shell will be the sa me everywhere as on its surface. As we add – 3Q charge on the surface, the potential on the surface changes by the same amount as that inside. Therefore the potential difference remains the same.

The equivalent capacitance

Q will remain same as no charge is leaving or entering the plates during the process of slab insertion

Now, Q = C ' V ' = C ' E 'd

Work done is the change in energy stored

The work done in moving a charge from A to B

W = (T.P.E.)_A – (T.P.E.)_B where T.P.E. = Total Potential Energy

Let λ be the charge per unit length. Let us consider a Gaussian surface (dotted cylinder).
Applying Gauss's law

For the flat portions of Gaussian surface, the angle between electric field and surface is 90°.
Hence flux through flat portions is zero.  

NOTE : By symmetry, the electric field on the curved surface is same throughout.

The angle between   (for curved surface)

The electric lines of force cannot enter the metallic sphere as electric field inside the solid metallic sphere is zero. Also, the origination and termination of the electric lines of force from the metallic surface is normally (directed towards the centre).

In region I and III, there will be electric field   directedfrom positive to negative. In region II, due to orientation of dipoles, there is an electric field  present in opposite
direction of   But since  is also present, the net electric field is  in the direction of    as shown in the diagram. (∵ E₀ > E_k)

NOTE : When one moves opposite to the direction of electric field, the potential always increases. The stronger the electric field, the more is the potential increase. Since in region II, the electric field is less as compared to I and III therefore the increase in potential will be less but there has to be increase in potential in all the regions from x = 0 to x = 3d. Also where

E is uniform, dV/dx = constt.

Potential at origin will be given by

Let Q be the charge on the ring, the negative charge – q is released from point P (0, 0, Z₀). The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be

Therefore, force on charge P will be towards centre as shown, and its magnitude is

Similarly, when it crosses the origin, the force is again towards centre O.
Thus the motion of the particle is periodic for all values of Z₀ lying between 0 and ∞.

Secondly if 

i.e. the restoring force F_e∝ – Z₀. Hence the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position)

KEY CONCEPT :  The expressions of the electric field inside the sphere   outside the sphere

Hence, E increases for r < R and decreases for R < r < ∞.

When two points are connected with a conducting path in electrostatic condition, then the potential of the two points is equal. Thus potential at A = Potential at B (c) is the correct option.
Option (d) is a result of Gauss's law

Total electric flux through cavity

Option (a) and (b) are dependent on the curvature which is different at points A and B.

(a) The  whole charge Q will be enclosed in a sphere of diameter 2R₀. (b) Electric field E = zero inside the sphere.
Hence electric field is discontinued at r = R₀.
(c) Changes in V and E are continuously present for r > R₀. Option (c) is incorrect.
(d) For r < R 0, t he potential V is constant and the electric intensity is zero.
Obviously, the electrostatic energy is zero for r < R₀.

The situation is shown in the figure which is similar to a planet revolving around sun. The distance of – q from + Q is changing, therefore, force between the charges will change.   

The speed of the charge – q will be greater when the charge is nearer to + Q as compared to when it is far.
Therefore,  the angular velocity of charge – q is also variable. The direction of the velocity changes continously, therefore, linear momentum is also variable.
The angular momentum of (– q) about Q is constant because the torque about + Q is zero.

The electric field lines are orginating from Q₁ and terminating on Q₂. Therefore Q₁ is positive and Q₂ is negative.
As the number of lines associated with Q₁ is greater than that associated with Q₂, therefore |Q₁| > |Q₂|.
Option (a) is correct.
At a finite distance on the left of Q₁, the electric field intensity cannot be zero because the electric field created by Q₁ will be greater than Q₂. This is because the magnitude of Q₁ is greater and the distance smaller 

At a finite distance to the right of Q₂, the electric field is zero. Here, the electric field created by Q₂ at a particular point will cancel out the electric field created by Q₁.

Electric field inside a spherical metallic shell with charge on the surface is always zero. Therefore option [a] is correct.
When the shells are connected with a thin metal wire then electric potentials will be equal, say V.

As R_A > R_B therefore Q_A > A_B. option [b] is also correct.

Option (d) is also correct

(a) is not correct because it is valid only when E ∝ r^–2
(b) is not correct
(c) is correct as between two point charges we will get a point where the electric field due to the two point charges cancel out each other.
(d) is correct when the work done is without accelerating the charge.

The electric flux passing through 

  is same due to symmetry.

The net electric flux through the cubical region is

∴  E_O = E_A + E_D + (E_F + E_C) cos 60º + (EB + EC) cos 60º

The electric potential at O is

PR is perpendicular bisector (the equatorial line) for the electric dipoles AB, FE and BC. Therefore the electric potential will be zero at any point on PR.
At any point ST, the electric field will be directed from S to T.
The potential decreases along the electric field line.

Electric field E₁ due to smaller sphere at P is

Electric field E₂ due to bigger sphere at P is

Option (d) is correct.

Step 1 : When S₁ is pressed. The capacitor C₁ gets charged such that its upper plate acquires a positive charge + 2 CV₀ and lower plate – 2 CV₀.
Step 2 : When S₂ is pressed (S₁ open). As C₁ = C₂ the charge gets distributed equal. The upper plates of C₁ and C₂ now take charge + CV₀ each and lower plate – CV₀ each.

(b) and (d) are correct option.

Let us consider a point P on the overlapping region. The electric field intensity at P due to positively charged sphere 

The electric field intensity at P due to negatively charged
sphere   The total electric field,

Therefore the electric field is same in magnitude and direction option (c) and (d) are correct.

∴ Q = 2λr₀                  ...(1)

This is a combination of two capacitors in parallel.
Therefore C = C₁ + C₂     ∴ C₂ = C – C₁

∴ (d) is a correct option.

     ∴ (c) is incorrect

Also V = E × d

   ∴ (a) is a correct option

Force on change q when it is given a small displacement x is F_net = F₁ – F₂

 and is directed towards the mean position therefore the charge +q will execute SHM.

In case of charge (–q) F₂ > F₁ therefore the charge –q continues to move in the direction of its displacement.

[C] is the correct option.

Assume the cavity to contain similar charge distribution of positive and negative charge as the rest of sphere.
Electric field at M due to uniformly distributed charge of the whole sphere of radius R₁

Electric field at M due the negative charge distribution in the cavity

∴ The total electric field at M is

(d) is the correct option