HC Verma Questions and Solutions: Chapter 30: Gauss’s Law- 1 | HC Verma Solutions - JEE PDF Download

The electric field remains same for the plastic plate and the copper plate, as both are considered to be infinite plane sheets. So, it does not matter whether the plate is conducting or non-conducting.
The electric field due to both the plates,

The particle is a conductor. When it is brought near a positively charged metal plate,  opposite charge is induced on its face nearer to the plate, i.e. negative and the same amount of charge, but of opposite polarity, goes to the farther end, i.e. positive.
Now, the attractive force is due to charges of opposite polarity. As they are at a lesser distance than the same polarity charges, the force of attraction is greater than the force of repulsion. In other words, the force on the particle is towards the plate.

A charge placed outside a conductor can induce a charge on it or can affect the charge on its surface. But it does not affect what is contained inside the conductor. Similarly, charge q₁ can affect charge Q; still, the force inside the conductor remains zero. An analogy to the above statement is that when lightning strikes a car, the charge that appears on the car's metal surface does not affect its interior. Due to this passengers are recommended to sit inside the car.

This question can be answered using the concept of electric field lines.
We know that electric field lines emerge from a positive charge and move towards a negative charge. Now, charge Q, on the face in front of charge q1, tries to nullify the field lines emerging from charge q1. So, the charge on the farther face of the conductor dominates and, hence, a force appears to the right of charge q.

The flux through a surface does not depend on its shape and size; it only depends upon the charge enclosed inside the volume. Here, the charge enclosed by the  sphere of radius 10 cm and the sphere of radius 20 cm is same so the flux through them will also be same.

At first, when the rod is inserted into the cube, the flux start increasing. When the rod is fully inserted, the flux becomes constant and remains constant for the remaining L/2 length of the rod. After that, as the rod moves out of the cube, the flux starts decreasing. These processes are depicted only by curve (d).

According to Gauss's Law, the flux through a closed cylindrical Gaussian surface is q/ε₀. But the question is about an open cylindrical vessel. Now, take another identical vessel and make a closed Gaussian surface enclosing the charge, as shown in the following  figure.

Total flux linked with the closed Gaussian surface,

Flux linked with the surface of a open ended cylindrical vessel,

The contribution of flux on the closed surface due to the charges lying outside the surface is zero because number of field line entering the closed surface is equal to the number of field lines coming out of the surface so the net contribution of the charge lying outside the closed surface to the flux is zero. Therefore, the net flux through the surface due to the charge lying outside the the closed surface is zero. The contribution that counts is only due to the charges lying within the closed surface.
Thus, the flux of the electric field through a closed surface due to all the charges (inside and outside the surface) is equal to the flux due to the charges enclosed by the surface.

Since the cube is metallic, the charge gets distributed on the surface and the interior remains charge-free. However, when a positive point charge Q is brought near the metallic cube, a negative charge gets induced on the face near the positive charge Q and an equal positive charge gets induced on the face, which is away from the charge Q. Thus, the metallic surface gets non-uniform charge distribution.

Since the non-conducting sheet M is given a uniform charge, it induces a charge in the metal rod A, which further induces a charge in the metal rod B, as shown in the figure. Hence, all the options are correct.

As the flux is zero through the surface, the charge enclosed must be zero. But the electric field is not necessarily zero everywhere on the surface. For example, in the case of a dipole enclosed by the surface, the electric field through the surface is not zero but has some value. So, the correct answers are (b) and (c).

The sphere encloses a dipole, i.e. two equal and opposite charges. In other words, net charge enclosed in the sphere is zero. Hence, the flux is zero through the sphere.
But the electric field at any point p on the sphere,
 
where θ is the angle made by the point p with the centre of the dipole.
Hence, we can see that the field is not zero anywhere on the sphere.

These are the only points in the straight line with the charge q and at the brim of the hemisphere. So, field lines emerging from these charges do not affect the flux through the hemisphere due to charge q. On the other hand, the two remaining charges are beside the surface of the hemisphere and not in line with the charge q and not at the brim. So, they will affect the flux. Hence, the correct answers are (a) and (c).

Initially, there is no charge in the closed surface. As the wire is neutral, the flux initially is zero. Now, if we connect the battery and a current flows through it, the flux remains zero, as the number of electrons entering the surface is equal to number of electrons leaving. That is, net charge enclosed is zero and so is the flux.

A positive charge at point P will induce a negative charge on the near face of the conducting sphere, whereas the positive charge on the farther end of the sphere. As this father end is enclosed or intersected by the closed surface, so the flux through it will become positive due to the induced positive charge.