__Gauss’ Law__

__Gauss’ Law__

The flux of the electric field from a volume is proportional to the charge inside—Gauss’ law—and the circulation of the electric field is zero—E is a gradient—are the two laws of electrostatics. All electrostatic predictions follow from these two principles. But it’s one thing to express these things analytically; it’s another to put them into practise quickly and with a certain bit of inventiveness. We’ll go through a few calculations that can be done directly with Gauss’ law in this chapter. We shall establish theorems and illustrate several consequences that may be easily understood from Gauss’ rule, particularly in conductors. Gauss’ law cannot solve any problem by itself since the other laws must also be followed.

So, if we want to apply Gauss’ law to solve a specific problem, we’ll have to add something to it. For example, we’ll need to assume some notion of how the field appears, based on symmetry considerations, for example. Alternatively, we may need to introduce the concept that the field is the gradient of a potential.

__Equilibrium In an Electrostatic Field__

__Equilibrium In an Electrostatic Field__

Take the following query first: When may a point charge in an electric field containing other charges be in stable mechanical equilibrium? Consider three negative charges at the corners of an equilateral triangle in a horizontal plane as an illustration. Would a positive charge placed in the triangle’s centre stay there? (It will be easier if we omit gravity for the time being; nonetheless, incorporating it would have no effect on the results.) Although there is no force on the positive charge, is the equilibrium stable? Is it possible that if the charge is slightly shifted, it will return to its equilibrium position? No, that is not the case.

In every electrostatic field, save exactly on top of another charge, there are no sites of stable equilibrium. It’s easy to understand why using Gauss’ law. To begin, the field must be zero for a charge to be in equilibrium at any given location P0. Second, if the equilibrium is to be maintained, there must be a restoring force directed opposite the displacement whenever the charge is moved away from P0 in any direction. All adjacent points’ electric fields must be oriented inward—towards point P0. But, as we can see, this is in violation of Gauss’ law if there is no charge at P0.

Consider a small imaginary surface enclosing P0. The surface integral of the normal component is not zero if the electric field everywhere in the vicinity is oriented toward P0. The flow through the surface must be negative in the instance depicted in the diagram. However, according to Gauss’ equation, the electric field flow through any surface is proportional to the total charge within.

The field we’ve imagined violates Gauss’ rule when there is no charges at P0. A positive charge cannot be balanced in empty space—at a position where there is no negative charge. If a positive charge is in the centre of a dispersed negative charge, it can be in equilibrium. Of course, something other than electrical forces would have to keep the negative charge distribution in place! For a point charge, we attained our outcome.

Is the same result applies to a complex arrangement of charges held together in fixed relative locations, such as with rods? For two equal charges mounted on a rod, we consider the problem. Is it feasible that under some electrostatic field, this combination will be in equilibrium? No, for displacements in all directions, the whole force on the rod cannot be restored.

Its total force just on rod at any position is denoted by F, which is a vector field. Following the logic of the preceding reasoning, we infer that the divergence of F must be negative at a stable equilibrium. However, the total force on the rod is equal to the first charge multiplied by the field at its current location, plus the second charge multiplied by the field at its current location: F=q_{1}E_{1}+q_{2}E_{2}.

The divergence of F is given by, ∇⋅F=q_{1}(∇⋅E_{1}) + q_{2}(∇⋅E_{2}). Both ∇⋅E_{1} and ∇⋅E_{2} are zero, and F is zero—not negative, as would be necessary for equilibrium—if each of the two charges q1 and q2 is in free space. An extension of the reasoning indicates that under an electrostatic field in free space, no rigid combination of any number of charges can have a stable equilibrium position.

We haven’t yet demonstrated that equilibrium is impossible in the presence of pivots or other mechanical limitations. Consider the case of a hollow tube in which a charge can easily move back and forth but not sideways. It is now quite simple to create an electric field that points inward at both ends of the tube provided the field is permitted to point laterally outward towards the tube’s centre. We simply insert positive charges at either end of the tube. Even if E’s divergence is zero, there can now be an equilibrium point. Without “nonelectrical” forces from the tube walls, the charge would not be in stable equilibrium during sideways motion.

Learn more about: a point charge q is placed at the centre of a cylinder of length L and radius R the electric flux through the curved surface of the cylinder is?