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a semi circular rod of radius R having a charge +Q distributed uniformly on it Linear charge density

By   April 17, 2023

note : Consider a semi-circular rod of radius R having a charge +Q distributed uniformly on it. Linear charge density is (Fig. below ).

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Consider a small element at A of length dl = Rdθ at A. The charge of the element is dQ = λdl = λRdθ. The electric field due to this element at centre C is

The x and y components of dE are dE cos θ and dE sin θ. Now consider an element at a symmetric point B. The x component of electric field of this element will cancel with that of element at A but y component will add up. Hence the electric field at C due to the complete semi-circular rod is

directed vertically downwards away from the rod. If the rod carries a negative charge (–Q), then the electric field is directed vertically upwards towards the rod.

The electric potential at point C is

The above results also hold for a non-conducting rod.

Coulomb’s Law

On the basis of his measurements, Coulomb arrived at a law, known after his name as Coulomb’s law, which states that the magnitude of the electric force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them, i.e

In the SI system, k is written as 1/4π ε0 where ε0 is called the permittivity of vacuum and its value is

The force F is attractive for unlike charges (q1 q2 < 0) and repulsive for like charges (q1 q2 > 0).

Coulomb’s Law in vector form

Case (a): Unlike charges (q1 q2 < 0) [Fig. above(a)] Force exerted on q2 by q1 is

Case (b): Like charges (q1 q2 > 0) [Fig. above (b)]

Relative Permittivity (or Dielectric Constant)

Relative permittivity of a medium is defined as the ratio of the permittivity of the medium to permittivity of vacuum, i.e.

εr is also called the dielectric constant (K) of the medium. Thus K = ε/ε0 or ε = Kε0. By definition K for air = 1. If charges q1 and q2 are situated in a medium other than air or vacuum, the magnitude of force between them is given by

Principle of Superposition

If many charges are present, the total force on a given charge is equal to the vector sum of the individual forces exerted on it by all other charges taken one at a time.

Electric Field

An electric field exists at any point in the space surrounding a charge. To define the electric field, we place a small positive point charge q0 at the point in space where the electric field is to be found and we measure the coulomb force F at that point. The electric field E is then given by

If a charge q is placed at a point where the electric field due to other charge or charges is E, then the charge q will experience a force F given by

F = qE

Electric field due to an isolated point charge

Electric field at a distance r from a source charge q is given by

For a positive charge (+q), vector E is directed radially outwards from it and for a negative charge (–q), E is directed radially inwards it. Because electric field E is vector quantity, the net electric field due to several charges is given by the vector sum of the electric fields due to the individual charges.

Electric field due to an electric dipole

A pair of equal and opposite point charges separated by a certain distance is called an electric dipole. Case (a): Electric field at a point on the axis of a dipole Let 2a be the separation between point charges –q and +q (Fig. below).

Electric fields at P due to +q and –q respectively are

where p = q(2a) is the dipole moment and 2a is the vector distance between charges –q and +q. Dipole moment p is a vector quantity directed from –q to +q. For a very short dipole (a << r)

Case (b): Electric field at a point on the perpendicular bisector (equatorial plane) of a dipole

Electric fields at point Q due to +q and –q are (see Fig. above)

note :  (i) The direction the electric field at a point on the axial line of a dipole is along the dipole moment.

(ii) The direction of the electric field at a point on the equatorial line of a dipole is antiparallel to the dipole moment.

Electric field due to a uniformly charged conducting rod

A conducting rod AB of negligible thickness and length L carries a charge a charge Q uniformly distributed on it. To find the electric field at point P at a distance a from end B (Fig. below), we consider a small element of length dx of the rod located at a distance x from end B.

Charge of element is dq = Q/ L dx = λ dx where λ = Q/ L is the linear charge density. The electric field due to the element at point P is

If Q is positive, E is directed from left to right.

Electric field due to a uniformly charged ring (or loop) of wire at a point on its axis

Consider a ring of radius R carrying a chare Q distributed uniformly on it. To find electric field at a point P on its axis at a distance x from the centre O, consider an element of length dl (Fig. below). The charge of the element is

The electric field due to element A is dE given by

There is a similar electric field at P due to element a diametrically opposite point B. The x components of electric fields due to these elements add up while the y components cancel. Hence

The direction of E is from O to P if charge Q is positive.

Electric Field Lines

Electric field lines of an electrostatic field give a pictorial representation of the field. An electric field line is a curve, the tangent to which at a point gives the direction of the electric field at that point. Figure below shows field line patterns around some charge distributions.

Properties of Electric Field Lines

(i) The tangent to a field line at any point gives the direction of electric field at that point.

(ii) Field lines originate from a positive charge and terminate on a negative charge.

(iii) No two field lines intersect.

(iv) Field lines are closer together in the region where the field is stronger and farther apart where the field is weaker.

(v) The number of field lines originating or ending on a charge is proportional to the magnitude of the charge.

note : Electric field line due to a charge distribution never forms a closed loop. But if the electric field is induced by a time-varying magnetic field, its field line forms a closed loop.

Electric Flux

The electric flux through a surface in an electric field is a measure of the number of electric field lines passing through the surface. For a plane surface of surface area S in an electric field E, the electric flux Φ is defined as

where S is called the area vector, its magnitude is S and its direction is normal to the surface and away from it. Angle θ is the angle between E and S. For a curved surface,

where n cap is a unit outward normal to the surface. dS is the surface area of an element of the surface (Fig. above). The SI unit of electric flux is NC–1 m2 or Vm (volt metre).

Gauss’s Law in Electrostatics

Gauss’s law states that the electric flux through a closed surface S in an electric field E is equal to q/ε0 , where q is the net charge enclosed in the surface and ε0 is electrical permittivity of vacuum.

Gauss’s law is used to obtain the expression for the electric field due to linear, surface and volume charge distributions which are uniform and symmetric so that a proper and convenient closed surface (called the Gaussian surface) can be chosen to evaluate the surface integral in Eq. (1).

Some Important Points about Gauss’s Law

(1) Gauss’s law holds for any closed surface of any shape or size.

(2) The surface that we choose to evaluate electric flux [i.e. to evaluate the surface integral in Eq. (1)] is called Gaussian surface.

(3) If the Gaussian surface is so chosen that there are some charges outside and some inside the surface, then charge q on the R.H.S. of Eq. (1) is the net charge (taking into account the sign of charges) enclosed inside the surface but electric field E on the L.H.S. of Eq. (1) is the electric field due to all the charges both inside and outside the surface.

(4) The exact location of charges inside Gaussian surface does not affect the value of the electric flux.

(5) If Coulomb’s law did not hold, Gauss’s law also would not hold.

Applications of Gauss’s Law

(1) Electric field due to a thin infinitely long straight charged rod or wire Electric field at a point at a perpendicular distance r from a thin, infinitely long straight rod or wire carrying a uniform linear charge density λ is given by

where λ = q /L is the charge per unit length of the rod and n cap is a unit vector pointing away from the rod if q is positive and towards it if q is negative.

(2) Electric field due to a thin sheet of charge

Electric field at a perpendicular distance r from a thin, flat and infinite sheet carrying a uniform surface charge density σ is given by

where σ = q / A is charge per unit area and n cap is a unit vector pointing away from the sheet if q is positive and towards it if q is negative. Notice that E is independent of r, the distance from the sheet.

Electric field due to a thin charged spherical shell

Electric field at a distance r from a spherical shell of radius R carrying a surface charge density σ (= q/4πR2 ) is given by

where n cap is a unit vector pointing radially outwards if q is positive and inwards if q is negative.

Electric Potential

The electric potential at a point in an electrostatic field is the work per unit charge that is done to bring a small charge in from infinity to that point along any path.

Electric potential due to an isolated point charge

Electric potential at a point P in the electric field of a point charge is given by

where r is the distance of the point P from the charge. This potential is spherically symmetric around the point, i.e. it depends only on r for a given charge q. Since potential is a scalar function, the spherical symmetry means that the potential at a point does not depend upon the direction of that point with respect to the point charge; it only depends on the distance of the point from the charge. Notice that the potential due to a positive charge (q > 0) is positive, it is negative in the neighborhood of an isolated negative charge (q < 0).

 Electric potential due to two point charges

To find the electric potential at a point in the electric field due to two or more charges, we first calculate the potential due to each charge, assuming that all other charges are absent, and then simply add these individual contributions. Since, unlike electric field, electric potential is a scalar, the addition here is the ordinary sum, not a vector sum. The potential at any point due to two point charges q1 and q2 is, therefore, simply given by

where r1 and r2 are the distances of the point in the question from charges q1 and q2 respectively.

Electric potential due to many point charges

The potential at any point due to a system of N point charges is given by

Relation between E and V

Electric field is the negative gradient of potential. This means that the potential decreases along the direction of the electric field.

Electric Potential Energy

The electric potential energy of a system of point charges is defined as the amount of work done to assemble this system of charges by bringing them in from an infinite distance. We assume that the charges were at rest when they

were infinitely separated, i.e. they had no initial kinetic energy. The electric potential energy of two point charges q1 and q2 separated by a distance r12 as shown in Fig. below (a) is given by

The electric potential energy of a system of three point charges as shown in Fig. below (b) is given by

This expression can be generalized for any number of charges.

The Electron-Volt T

he SI unit of potential energy is the joule. In atomic physics a more convenient unit called the electron-volt (written as eV) is used. An electron-volt is the potential energy gained or lost by an electron in moving through a potential difference of 1 volt. Since the magnitude of charge on an electron is 1.6 x 10–19 C, 1 eV = 1.6  x 10–19 J

 Potential Energy of an Electric Dipole in an External Electric Field

When a dipole is placed in a uniform electric field E, as shown in Fig. below, it experiences a torque given by

τ = p E sinθ

where θ is the angle between the line joining the two charges and the electric field. In vector form

τ = p x E

The torque tends to rotate the dipole to a position where θ = 0, i.e, p is parallel to E.

The electric potential energy of a dipole is U = – p . E

Additional Useful Formulae

(1) Electric field and potential due to a group of charges

(i) Charge q at each vertex of an equilateral triangle of side a (Fig. below).

(ii) Charge q at each vertex of a square of side a (Fig. below).

note : In the above two cases, if one of the charges is removed from a vertex, the net electric field at O and C is E = q/4πε0 r 2 , directed towards the empty vertex