Electrostatic Force, Field and Potential Flipbook PDF

Electrostatic Force, Field and Potential

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Electric Field Intensity and Electric Flux

Coloumb's Law Coloumb's Law is applied to calculate the force of attraction or repulsion between two-point charges. It states that "the force of attraction or repulsion between two charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them."

Figure: Two charges separated by a distance r apart. Mathematically, $$F\propto q_1q_2\dots(i)$$ $$F\propto \frac{1}{r^2}\dots(ii)$$ Combining equation (i) and (ii), we get $$F\propto \frac{q_1q_2}{r^2}$$ $$F=K \frac{q_1q_2}{r^2}$$ Here, K is the proportionality constant whose value depends upon the medium in which charges are present and the system of a unit chosen. For S.I. system and charges present in the air or Vaccum $$K = \frac{1}{4\pi \epsilon _o}(\text{ Here} \epsilon _o \text {is permitivity of vaccum or free space.})$$ $$\therefore F = \frac{1}{4\pi \epsilon _o } \frac{q_1q_2}{r^2}$$ In C.G.S K = 1 $$\therefore F=\frac{q_1q_2}{r^2}$$ Relative Permittivity \((\epsilon _r)\) When two charges \(q_1 \text{and} q_2\) are at the distance'r' apart from each other in vacuum \(\epsilon _o\) the force is given by $$F_v = \frac{1}{4\pi \epsilon _o } \frac{q_1q_2}{r^2}\dots(i)$$ When the charges are placed at the same distance in a medium having permittivity \(' \epsilon '\) then the force is $$ F_m = \frac{1}{4\pi \epsilon _o } \frac{q_1q_2}{r^2}$$

Now dividing equation (i) by (ii), we get $$\frac {F_v = \frac{1}{4\pi \epsilon _o } \frac{q_1q_2}{r^2}}{F_m = \frac{1}{4\pi \epsilon _o } \frac{q_1q_2}{r^2}}$$ $$\frac{F_v}{F_m} = \frac{\epsilon }{\epsilon _o} = \epsilon _r$$ Hence, the dielectric constant or relative permittivity of a medium can be defined as the ratio of the permittivity of a medium and the permittivity of vacuum of free space. In terms of force between charges, the dielectric constant of a medium can be defined as the ratio between two charges at certain distance in vacuum and the force between the same charges placed at the same distance is such medium. Electric Field The region around a charge where its electrostatic force of attraction or repulsion can be experienced is called an electric field of that charge. Outside the electric field of a charge, its influence is 0. Electric Field Intensity The electric field intensity of a point inside the electric field of a charge is the force experienced by a unit positive charge (+1 coulomb) placed at that point. It is a vector quantity having unit N/C.

Fig: Electric field intensity due to a point charge. To calculate electric field intensity(E) at a point 'p' near the charge (+q o) at point (p). The force experienced by the charge 'qo' is $$ F = \frac{1}{4\pi \epsilon _o } \frac{qq_o}{r^2}$$ So, by the definition of electric field intensity(E) $$E = \frac{f}{q_o}$$ $$ =\frac{f}{q_o} \left ( \frac{1}{4\pi \epsilon _o } \frac{qq_o}{r^2} \right )$$ $$\therefore E = \frac{1}{4\pi \epsilon _o } \frac{q}{r^2}$$ This is the electric field intensity produced by a point charge 'q' at distance ''r from it. Electric Field Intensity due to Number of Charge The electric field intensity of a point due to a number of a charge is equal to the vector sum of an electric field intensity of individual charge. If \(\vec E_1, \vec E_2, \vec E_3, \dots , \vec E_n\) be the electric field intensity at a point due to different charges then the net electric field intensity\(\vec E\) at that point is give by $$\vec E = \vec E_1, \vec E_2, \vec E_3, \dots , \vec E_n $$ Electric Lines of Force

The electric Lines of force are the imaginary lines in an electric field such that a tangent drawn at any point on it gives the direction of an electric field intensity at that point. The number of electric lines of force crossing an area gives the measure of electric field intensity. Properties of Electric Lines of Force 1. 2. 3. 4. 5.

Electric lines of force originated from a positive charge and terminate at a negative charge. They tend to contract longitudinally because of attraction between unlike charges, They exert lateral pressure on each other because of repulsion between like charges. Electric lines of force are continuous curves. A tangent drawn at a point in electric lines of force gives the direction of electric field intensity at that point. 6. Two electric lines of force never intercept each other. 7. Electric lines of force leave the surface perpendicularly.

flux through the surface area Electric Flux \(\phi\) The number of electric lines of force passing through an area held perpendicularly is called electric flux. Larger the value of electric flux greater will be the electric field intensity. Electric field intensity can be defined as the electric flux passing through unit area held perpendicularly. i.e. $$\text{Electric field intensity (E)} = \frac{Electric flux(\phi)}{Area(A)}$$ $$E = \frac{\phi}{A}$$ $$\therefore \phi = EA In vector form

$$\phi = \vec E. \vec A$$ Hence, electric lines flux can be defined as the scalar product of electric flux intensity and vector area. Electric Dipole and Dipole Two equal and opposite charges separated at certain finite distance constitutes electric dipole. The dipole moment of an electric dipole is defined as the product of the two equal charges and perpendicular distance between them i.e. dipole moment \((\vec p) = q\vec d\). The dipole moment of an isolated atom is zero because the centre of positive and negative charge coincides. The dipole moment exists only when the positive and negative centres are separate.

Gauss's Law and it's Application Gauss's Law

Statement The total dielectric flux passing through a closed surface in vacuum enclosing a charge is \(\frac{1}{\epsilon _o}\) times the charge enclosed by the closed surface. Mathematically, $$\phi = \frac{1}{\epsilon _o} \times \text{charge enclosed (q)} $$ $$\phi = \frac{q}{\epsilon _o}$$ where,\(\epsilon _o\) is permittivity of free space. In medium,$$\phi = \frac{q}{\epsilon}$$ To verify Gauss theorem suppose a point charge placed at O in vacuum show that the electric field intensity at point 'P' that lies at distance'r' from the charge 'q' is $$E = \frac{1}{4\pi \epsilon _o}. \frac{q}{r^2}$$ When a spherical surface passing through point'P' is constructed that has the centre 'O' and radius 'r' at every pointon its surface electric field intensity has the same value. If \('\phi '\) represents the electric flux passing through the sphere and 'A' its area then $$\phi = EA$$ $$= \frac{1}{4\pi \epsilon _o}. \frac{q}{r^2} 4\pi r^2$$ $$=\frac{q}{\epsilon _o}$$ $$\therefore \phi=\frac{q}{\epsilon _o}$$ Application of Gauss's Theorem Gauss's theorem is useful for determining the electric field intensity produced by a charged conductor as follows: Electric Field Intensity due to a Charged Sphere

Consider a sphericalConductor having radius 'R' and center 'O'. It is carrying charge 'q' on it's surface. Due to this distribution of charged electric field intensity around it can be defined. Following three distinct points can be considered.

Point P lying outside the charged sphere 1. At a point outside the surface of sphere(r>R) Consider 'P' is the concerned point where electric field intensity due to the distribution of charge on the spherical conductor is to be calculated. For this, we construct a Gauss' sphere concentric with the charged sphere and passing through point 'P'. So, that due to symmetry at every point on the surface of the sphere electric field intensity has the same magnitude. It \(\phi\) represents the electric flux passing through Gauss' surface and 'A' be area of Gauss' surface and 'A' be area of Gauss's Sphere. Then $$\phi = EA$$ $$\phi= E4\pi r^2 \dots(i)$$ Also from Gauss' theorem $$\phi =\frac{ \text{charge enclosed by Gaussian Surface}}{\epsilon _o}$$ $$\phi =\frac{q}{\epsilon _o} \dots (ii)$$ Equating (i) and (ii), we get $$E. 4\pi r^2 = \frac{q}{\epsilon _o}$$ $$\therefore E = \frac{1}{4\pi r^2}.\frac{q}{\epsilon _o}$$ This is the electric field intensity produced distanced 'r' from the center of a spherical conductor carrying charge 'q'.

Point P lying on the surface of the sphere. 2. At a point on the surface of sphere(r=R) When the concerned point lies on the surface the gauss' surface also becomes the sphere having same radius as that of the charged surface. Then, $$\phi = EA$$ $$\phi= E4\pi r^2 \dots(i)$$ Also from Gauss' theorem $$\phi =\frac{ \text{charge enclosed by Gaussian Surface}}{\epsilon _o}$$ $$\phi =\frac{q}{\epsilon _o} \dots (ii)$$ Equating (i) and (ii), we get $$E. 4\pi R^2 = \frac{q}{\epsilon _o}$$ $$\therefore E = \frac{1}{4\pi \epsilon}.{\frac{q}{R^2}}$$ Hence, electric field intensity on the surface of charged sphere has constant magnitude. $$E = \frac{1}{4\pi r^2}. \frac{q}{\epsilon _o}$$

Point P lying inside the sphere. 3. At a point inside the surface of sphere(r