Physics Electrostatics Class 12 Notes 2 Learn CBSE are prepared for CBSE and RBSE students both. These notes are so much useful for the board exam. These notes are in a short form for your exam point with all derivations. Physics Notes Class 12. electric charges and field notes. Electrostatics Class 12 Notes The notes are available in pdf form.
Electrostatic Potential and Potential Difference 1
CONTINUOUS CHARGE DISTRIBUTION
Continuous charge distribution. In practice, we deal with charges much greater in magnitude than the charge on an electron, so we can ignore the quantum nature of charges and imagine that the charge is spread in a region in a continuous manner. Such a charge distribution is known as a continuous charge distribution.
Calculation of the force on a charge due to a continuous charge distribution. As shown in Fig. 1.57, consider a point charge q0 lying near a region of continuous charge distribution. This continuous charge distribution can be imagined to consist of a large number of small charges dq. According to Coulomb’s law, the force on point charge q0 due to small charge dq is
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where r^=r^/r , is a unit vector pointing from the small charge dq towards the point charge q0. By the principle of superposition, the total force on charge q0 will be the vector sum of the forces exerted by all such small charges and is given by
Fig Force on a point charge q 0 due to a continuous charge distribution.
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Name the different types of continuous charge distributions. Define their respective charge densities. Write expression for the electric field produced by each type of charge distribution. Hence write expression for the electric field of a general source charge distribution.electric charges and field notes
Different types of continuous charge distributions. There are three types of continuous charge distributions: ELECTROSTATICS Notes
(a) Volume charge distribution. It is a charge distribution spread over a three dimensional volume or region V of space, as shown in Fig. 1.57. We define the volume charge density at any point in this volume as the charge contained per unit volume at that point, i.e.,
r =
The SI unit for r is coulomb per cubic meter (Cm 3).
For example, if a charge q is distributed over the entire volume of a sphere of radius R, then its volume charge density is
r =
Fig. Volume charge distribution
The charge contained in small volume dV is
dq = r dV
Total electrostatic force exerted on charge q0 due to the entire volume V is given by
Electric field due to the volume charge distribution at the location of charge q0 is
(b) Surface charge distribution. It is a charge distribution spread over a two-dimensional surface S in space, as shown in Fig. 1.59. We define the surface charge density at any point on this surface as the charge per unit area at that point, i.e.,
s =
The SI unit for σ is Cm-2.
Fig. Surface charge distribution.
For example, if a charge q is uniformly distributed over the surface of a spherical conductor of radius R, then its surface charge density is
s = Cm-2
The charge contained in small area dS is
dq = σ dS
Total electrostatic force exerted on charge q0 due to the entire surface S is given by
Electric field due to the surface charge distribution at the location of charge q0 is
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(c) Line charge distribution. It is a charge distribution along a one-dimensional curve or line L in space, as shown in Fig. 1.60. We define the line charge density at any point on this line as the charge per unit length of the line at that point, i.e.,
l =
The SI unit for l is Cm-1.
Fig. Line charge distribution.
For example, if a charge q is uniformly distributed over a ring of radius R, then its linear charge density is
Total electrostatic force exerted on charge due to the entire length is given by
Formulae Used
1. Volume charge density, =
2. Surface charge density, σ =
3. Linear charge density, l =
4. Force exerted on a charge q0 due to a continuous charge distribution,
5. Electric field due to a continuous charge distribution,
Units Used
is in Cm-3, σ in Cm-2, l in Cm_1 and E in NC_1.
AREA VECTOR
We come across many situations where we need to know not only the magnitude of a surface area but also its direction. The direction of a planar area vector is specified by the normal to the plane. In Fig.(a), a planar area element dS has been represented by a normal vector . The length of vector represents the magnitude dS of the area element. If is unit vector along the normal to the planar area, then
= dS
Fig.(a) A planar area element. (b) An area element of a curved surface.
In case of a curved surface, we can imagine it to be divided into a large number of very small area elements. Each small area element of the curved surface can be treated as a planar area. By convention, the direction of the vector associated with every area element of a closed surface is along the outward drawn normal. As shown in Fig. 1.81(b), the area element at any point on the closed surface is equal to dS , where dS is the magnitude of the area element and is a unit vector in the direction of outward normal. CBSE
ELECTRIC FLUX
The term flux implies some kind of flow. Flux is the property of any vector field. The electric flux is a property of electric field.
The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area. Physics Notes For Class 12 Chapter-1 part 2-Learn CBSE
As shown in Fig. if an electric field passes normally through an area element ∆S, then the electric flux through this area is
∆ϕE = E ∆S
Fig. Electric flux through normal area.
As shown in Fig. if the normal drawn to the area element ∆S makes an angle θ with the uniform
Fig. Flux through an inclined area.
In case the field is non-uniform, we consider a closed surface S lying inside the field, as shown in Fig. We can divide the surface S into small area elements : Δ 2, ,…, . Let the corresponding electric fields at these elements be
Fig. Electric flux through a closed surface S.
Then the electric flux through the surface S will be
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Thus the electric flux through any surface , open or closed, is equal to the surface integral of the electric field taken over the surface .
Electric flux is a scalar quantity.
Unit of = Unit of E × unit of S
∴ SI unit of electric flux
= NC-1.m2 = Nm2C_1.
Equivalently, SI unit of electric flux
= Vm_1.m2 = Vm.
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GAUSS’S THEOREM
This theorem gives a relationship between the total flux passing through any closed surface and the net charge enclosed within the surface.
Gauss theorem states that the total flux through a closed surface is 1/ε0 times the net charge enclosed by the closed surface.
Mathematically, it can be expressed as
Proof. For the sake of simplicity, we prove Gauss’s theorem for an isolated positive point charge q. As shown in Fig. 1.85, suppose the surface S is a sphere of radius r centered on q. Then surface S is a Gaussian surface.
Fig. Flux through a sphere enclosing a point charge.
Electric field at any point on S is
This field points radially outward at all points on S. Also, any area element points radially outwards, so it is parallel to , i.e., θ = 0°.
∴ Flux through area is
d = . = E dS cos 0° = EdS
Total flux through surface S is
Total area of sphere
or
This proves Gauss’s theorem. CBSE
- Gauss’s theorem is valid for a closed surface of any shape and for any general charge distribution.
- If the net charge enclosed by a closed surface is zero (q = 0), then flux through it is also zero.
- The net flux through a closed surface due to a charge lying outside the closed surface is zero.
- The charge q appearing in the Gauss’s theorem includes the sum of all the charges located anywhere inside the closed surface.
- The electric field appearing in Gauss’s theorem is due to all the charges, both inside and outside the closed surface. However, the charge q appearing in the theorem is only contained within the closed surface.
- Gauss’s theorem is based on the inverse square dependence on distance contained in the coulomb’s law. In fact, it is applicable to any field obeying inverse square law. It will not hold in case of any departure from inverse square law.
- For a medium of absolute permittivity ε or dielectric constant , the Gauss’s theorem can be expressed as
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GAUSSIAN SURFACE
Any hypothetical closed surface enclosing a charge is called the Gaussian surface of that charge. It is chosen to evaluate the surface integral of the electric field produced by the charge enclosed by it, which, in turn, gives the total flux through the surface.
Importance. By a clever choice of Gaussian surface, we can easily find the electric fields produced by certain symmetric charge configurations which are otherwise quite difficult to evaluate by the direct application of Coulomb’s law and the principle of superposition.
COULOMB’S LAW FROM GAUSS’S THEOREM
Deduction of Coulomb’s law from Gauss’s theorem. As shown in Fig. consider an isolated positive point charge q. We select a spherical surface S of radius r centered at charge q as the Gaussian surface.
Fig. Applying Gauss’s theorem to a point charge.
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Formulae Used
1. Electric flux through a plane surface area S held in a uniform electric field is
ϕE = E. S= ES cos θ
where θ is the angle which the normal to the outward drawn normal to surface area makes with the field .
2. According to Gauss’s theorem, the total electric flux through a closed surface S enclosing charge q is
Units Used
Electric flux ϕE is in Nm 2 C-1 and flux density in NC-1.
Constant Used
Permittivity constant of free space is
C2 N_1m-2
ELECTROSTATICS Notes
APPLICATIONS OF GAUSS THEOREM
1.FIELD DUE TO AN INFINITELY LONG CHARGED WIRE
Electric field due to an infinitely long straight charged wire. Consider a thin infinitely long straight wire having a uniform linear charge density l Cm-1. By symmetry, the field of the line charge is directed radially outwards and its magnitude is same at all points equidistant from the line charge. To determine the field at a distance r from the line charge, we choose a cylindrical Gaussian surface of radius r, length l and with its axis along the line charge. As shown in Fig. it has curved surface S1 and flat circular ends S2 and S3. Obviously, 1 || , 2 ^ and 3 ^ . So only the curved surface contributes towards the total flux.
Fig. Cylindrical Gaussian surface for line charge.
Thus the electric field of a line charge is inversely proportional to the distance from the line charge.
2.ELECTRIC FIELD DUE TO A UNIFORMLY CHARGED INFINITE PLANE SHEET
Electric field due to a uniformly charged infinite plane sheet. As shown in consider a thin, infinite plane sheet of charge with uniform surface charge density σ. We wish to calculate its electric field at a point P at distance r from it.
Fig. Gaussian surface for a uniformly charged infinite plane sheet.
By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P and F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross- sectional area A and length 2r with its axis perpendicular to the sheet.
As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is
= EA+ EA=2 EA
Charge enclosed by the Gaussian surface,
Q = σA
According to Gauss’s theorem,
∴ 2EA = or E =
Clearly, E is independent of r, the distance from the plane sheet.
(i) If the sheet is positively charged (σ > 0), the field is directed away from it.
(ii) If the sheet is negatively charged (σ < 0), the field is directed towards it.
For a finite large planar sheet, the above formula will be approximately valid in the middle regions of the sheet, away from its edges. CBSE
Two infinite parallel planes have uniform charge densities of σ1 and σ2. Determine the electric field at points (i) to the left of the sheets, (ii) between them, and (iii) to the right of the sheets.
Electric field of two positively charged parallel plates. Fig. 1.99 show’s two thin plane parallel sheets of charge having uniform charge densities σ1 and σ2 with σ1 > σ2 > 0. Suppose is a unit vector pointing from left to right.
Fig.
In the region I : Fields due to the two sheets are
In the region II: Fields due to the two sheets are
∴ Total field,
In the region III: Fields due to the two sheets are
∴ Total field,
Two infinite parallel planes have uniform charge densities ± σ. Determine the electric field in (i) the region between the planes, and (ii) outside it.
Electric field of two oppositely charged plane parallel plates. As shown in Fig. consider two plane parallel sheets having uniform surface charge densities of ± σ. Suppose be a unit vector pointing from left to right.
Fig.
In the region I : Fields due to the two sheets are
Total field,
In the region II: Fields due to the two sheets are
∴ Total field,
In the region III: Fields due to the two sheets are
∴ Total field,
Thus the electric field between two oppositely charged plates of equal charge density is uniform which is equal to and is directed from the positive to the negative plate, while the field is zero on the outside of the two sheets. This arrangement is used for producing uniform electric field.
3.FIELD DUE TO A UNIFORMLY CHARGED THIN SPHERICAL SHELL
Apply Gauss’s theorem to show that for a spherical shell, the electric field inside the shell vanishes, whereas outside it, the field is as if all the charge had been concentrated at the center. CBSE
Electric field due to a uniformly charged thin spherical shell. Consider a thin spherical shell of charge of radius R with uniform surface charge density σ. From symmetry, we see that the electric field at any point is radial and has same magnitude at points equidistant from the centre of the shell i.e., the field is spherically symmetric. To determine electric field at any point P at a distance r from O, we choose a concentric sphere of radius r as the Gaussian surface.
Fig. Gaussian surface for outside points of a thin spherical shell of charge.
(a) When point P lies outside the spherical shell. The total charge q inside the Gaussian surface is the charge on the shell of radius R and area 4πR2.
∴ q = 4πR2 σ
Flux through the Gaussian surface, ϕE = E × 4 πr2
By Gauss’s theorem,
ϕE =
∴ E × 4πr2 =
or E = [For r > R]
This field is the same as that produced by a charge q placed at the centre O. Hence for points outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre. electric charges and field notes
(b) When point P lies on the spherical shell. The Gaussian surface just encloses the charged spherical shell.
Applying Gauss’s theorem,
E × 4πR2 =
or E = [For r = R]
or E = [∵ q = 4πR2 σ]
(c) When point P lies inside the spherical shell. As is clear from Fig. 1.102(a), the charge enclosed by the Gaussian surface is zero, i.e.,
q = 0
Fig. Gaussian surface for inside points of a thin spherical shell of charge.
Flux through the Gaussian surface,
ϕE = E × 4πr2
Applying Gauss’s theorem,
E × 4πr2 = 0
or E = 0 [For r < R]
Hence electric field due to a uniformly charged spherical shell is zero at all points inside the shell.
Figure shows how E varies with distance r from the centre of the shell of radius R. E is zero from r = 0 to r = R ; and beyond r = R, we have
Fig. Variation of E with r for a spherical shell of charge.
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4.FIELD DUE TO A UNIFORMLY CHARGED INSULATING SPHERE cbse
A charge q is uniformly distributed within an insulating sphere of radius R. Apply Gauss’s theorem to find the electric field due to this charge distribution at a point distant r from the centre of the sphere, where (a) r > R (b) r = R (c) 0 < r < R. Show the variation of E with r graphically.
Electric field due to uniformly charged insulating sphere. Consider an insulating sphere (For example, a nucleus with protons almost uniformly distributed inside it) of radius A with uniform volume charge density p. From symmetry, we see that the electric field at any point is radial and has same magnitude at points equidistant from the center O of the sphere i.e., the field is spherically symmetric. To determine electric field at any point P at distance r from O, we choose a concentric sphere of radius r as the Gaussian surface.
(a) When point P lies outside the sphere. The total charge q inside the Gaussian surface is the charge inside the sphere the radius R.
∴
Flux through the Gaussian surface,
ϕE = E × 4 πr2
Fig. Gaussian surface for outside points of an insulating sphere of charge.
By Gauss’s theorem,
E × 4πr2 =
or E = [For r > R]
This field is same as that produced by a charge q placed at the centre O. Hence for points outside the sphere the field due to the uniformly charged sphere is as if the entire charge of the sphere is concentrated at its centre.
(b) When the point P lies on the sphere. The Gaussian surface just encloses the charged sphere. Applying Gauss’s theorem,
E × 4πR2 =
Or [For r = R]
(c) When point P lies inside the sphere. The charge enclosed by the Gaussian surface is
Fig. Gaussian surface for inside points of an insulating sphere of charge.
By Gauss’s theorem,
E × 4πr2 =
or E × 4πr2 =
or E =
At the center, r = 0 and hence E = 0. The variation of E with distance r from the center of a sphere of uniformly charged sphere is shown in the figure 1.103(c).
Fig. Variation of E with r for a uniformly charged sphere.
Formulae Used
1. Electric field of a long straight wire of uniform linear charge density <lemma>,
E =
where r is the perpendicular distance of the observation point from the wire.
2. Electric field of an infinite plane sheet of uniform surface charge density σ,
E =
3. Electric field of two positively charged parallel plates with charge densities σ1 and σ2 such that σ1 > σ2 > 0.
E = ± (σ1 + σ2) (Outside the plates)
E = (σ1 – σ2) (Inside the plates)
4. Electric field of two equally and oppositely charged parallel plates,
E= 0 (For outside points)
E = (For inside points)
5. Electric field of a thin spherical shell of charge density σ and radius R,
E = For r > R (Outside points)
E= 0 For r < R (Inside points)
E = For r = R (At the surface)
Here q = 4π R2 σ.
6. Electric field of a solid sphere of uniform charge density p and radius R :
E = For r > R (Outside points)
E = For r < R (Inside points)
E = For r = R (At the surface)
Here q = π R3r
Units Used
Here charges are in coulomb, r and R in meter, l in Cm-1, σ in Cm-2, r in Cm-3 and electric field E in NC-1 or Vm-1.