Electrostatic Force and Electrostatic field:- |
Electrostatic:- It is a branch of physics that deals with the phenomena and properties of stationary or slow-moving electric charges with no acceleration.
F = Kq1q2/r2
Here, K = 1/4πε0 = 9×109 Nm2C-2 (in free space)
The relative permittivity (εr) of a medium is defined as the ratio between its permittivity of the medium (ε) and the permittivity (ε0) of the free space.
εr = ε/ε0
![\vec{F_{12}} = [q_{1}q_{2}/r^{2}]\hat{r_{21}}](https://files.askiitians.com/cdn1/cms-content/common/latex.codecogs.comgif.latex_vecf_12q_1q_2r2_hatr_21.jpg)
If q1q2>0, R.H.S is positive.
If q1q2<0, a negative sign from q1q2 will change
and
. The relation will again be true, since, in that case have same directions.
C.G.S, q = ±1 stat-coulomb
S.I, q = ±1 Coulomb
1 coulomb = 3×109 stat-coulomb
1 coulomb =(1/10) ab-coulomb (e.m.u of charge)
So, εr = ε/ε0 = F1/F2
Here, F1 and F2 are the magnitudes of the force between them in free space and in a medium respectively.
? Line charge, λ = q/L
Surface charge, σ = q/A
Volume charge, ρ = q/V
) :- The strength of an electric field is measured by the force experienced by a unit positive charge placed at that point. The direction of field is given by the direction of motion of a unit positive charge if it were free to move. 
E = [Newton/Coulomb] or [Joule/(Coulomb) (meter)]

Properties:-
(a) The lines of force are directed away from a positively charged conductor and are directed towards a negatively charged conductor.
(b) A line of force starts from a positive charge and ends on a negative charge. This signifies line of force starts from higher potential and ends on lower potential.
?
(a) At point on its axis.
E = (λ/4πε0) [1/a – 1/a+L]
Here, λ is the linear charge density.
(b) At a point on the line perpendicular to one end.
?
Here λ is the line charge.

At a point on its axis, E = (1/4πε0) [qx/(a2+x2)3/2]

Here σ is the surface charge.
(a) Eout = (1/4πε0) (q/r2)
(b) Ein = 0


(a) Outside Point:- Eout = (1/4πε0) (Q/r2)
(b) Inside Point:- Ein = (1/4πε0) (Qr/R3)
(c) On the Surface:- Esurface = (1/4πε0) (Q/R2)
Here, Q is the total charge

(a) Outside the cylinder:- E = λ/2πε0r
(b) Inside the cylinder:- E = 0
(a) Outside the cylinder:- E = λ/2πε0r
(b) Inside a point:- E = ρr/2ε0

(a) Electric field at points outside the charged sheets:-
EP = ER = 0
(b) Electric field at point in between the charged sheets:-
EQ = σ/ε0
) of an electric dipole is defined as the product of the magnitude of one of the charges and the vector distance from negative to positive charge. 
(a) At any point on the axial line:-

Alt Tag: Electric field due to an electric dipole on the axial line.
(b) At a point on the equatorial line (perpendicular bisector):-


(c) At any point:-


) acting on a electric dipole in a uniform electric field (E):-
= pE sinθ
Here, p is the dipole moment and θ is the angle between direction of dipole moment and electric field E.
and
for all the elementary areas constituting the surface. 

E = λ/2πε0r
The direction of electric field E is radially outward for a line of positive charge.

(a) Point at outside (r > R):- E = (1/4πε0) (q/r2), Here q is the total charge.
(b) Point at inside (r < R):- E = (1/4πε0) (qr/R3), Here q is the total charge.
E = σ/2ε0

This signifies, the electric field near a charged sheet is independent of the distance of the point from the sheet and depends only upon its charge density and is directed normally to the sheet.
?E= σ/ε0
Pelec = (½ε0) σ2
Electro-Static Potential and Capacitance:- |
(a) Electric potential, at any point, is defined as the negative line integral of electric field from infinity to that point along any path.

(b) V(r) = kq/r
(c) Potential difference, between any two points, in an electric field is defined as the work done in taking a unit positive charge from one point to the other against the electric field.
WAB = q [VA-VB]
So, V = [VA-VB] = W/q
Units:- volt (S.I), stat-volt (C.G.S)
Dimension:- [V] = [ML2T-3A-1]
Relation between volt and stat-volt:- 1 volt = (1/300) stat-volt
E = -dV/dx = --dV/dr
V = (1/4π ε0) (q/r)
V = (1/4π ε0) [q1/r1 + q2/r2 + q3/r3]
= V1+V2+ V2+….
(a) Outside, Vout = (1/4π ε0) (q/r)
(b) Inside, Vin = - (1/4π ε0) (q/R)
(c) On the surface, Vsurface = (1/4π ε0) (q/R)
(a) Outside, Vout = (1/4π ε0) (q/r)
(b) Inside, Vin = (1/4π ε0) [q(3R2-r2)/2R3]
(c) On the surface, Vsurface = (1/4π ε0) (q/R)
(d) In center, Vcenter = (3/2) [(1/4π ε0) (q/R)] = 3/2 [Vsurface]
(a) Common potential, V = (1/4π ε0) [(Q1+Q2)/(r1+r2)]
(b) q1 = r1(Q1+Q2)/(r1+r2) = r1Q/ r1+r2 ; q2 = r2Q/ r1+r2
(c) q1/q2 = r1/r2 or σ1/ σ2 = r1/r2
V (r,θ) = qa cosθ/4πε0r2 = p cosθ/4πε0r2
(a) Point lying on the axial line:- V = p/4πε0r2
(b) Point situated on equatorial lines:- V = 0
(a) R = n1/3r
(b) Q = nq
(c) V = n2/3Vsmall
(d) σ = n1/3 σsmall
(e) E = n1/3 Esmall
W = U = (1/4πε0) (q1q2/r12) = q1V1
W = U = (1/4πε0) (q1q2/r12 + q1q3/r13 + q2q3/r23)

(a) If θ = 90º, then W = 0
(b) If θ = 0º, then W = -pE
(c) If θ = 180º, then W = pE
K. E = ½ mv2 = eV
Conductors:- Conductors are those substance through which electric charge easily.
Insulators:- Insulators (also called dielectrics) are those substances through which electric charge cannot pass easily.
Capacity:- The capacity of a conductor is defined as the ratio between the charge of the conductor to its potential
C = Q/V
Units:-
S.I – farad (coulomb/volt)
C.G.S – stat farad (stat-coulomb/stat-volt)
Dimension of C:- [M-1L-2T4A2]
C = 4πε0r
Capacitor:- A capacitor or a condenser is an arrangement which provides a larger capacity in a smaller space.
Capacity of a parallel plate capacitor:-
Cair = ε0A/d
Cmed = Kε0A/d
Here, A is the common area of the two plates and d is the distance between the plates.
C = ε0A/[d-t+(t/K)]
Here d is the separation between the plates, t is the thickness of the dielectric slab A is the area and K is the dielectric constant of the material of the slab.
If the space is completely filled with dielectric medium (t=d), then,
C = ε0KA/ d
(a) Cair = 4πε0R
(b) Cmed = K (4πε0R)
(a) When outer sphere is earthed:-
Cair = 4πε0 [ab/(b-a)]
Cmed = 4πε0 [Kab/(b-a)]
(b) When the inner sphere is earthed:-
C1= 4πε0 [ab/(b-a)]
C2 = 4πε0b?
Net Capacity, C '=4πε0[b2/b-a]
Increase in capacity, ΔC = 4π ε0b
It signifies, by connecting the inner sphere to earth and charging the outer one we get an additional capacity equal to the capacity of outer sphere.
Cair = λl / [(λ/2π ε0) (loge b/a)] = [2π ε0l /(loge b/a) ]
Cmed = [2πKε0l /(loge b/a) ]
W = ½ QV = ½ Q2/C = ½ CV2
U = ½ ε0E2 = ½ (σ2/ ε0)
This signifies the energy density of a capacitor is independent of the area of plates of distance between them so long the value of E does not change.
?(a)
(i) Capacitors in parallel:- C = C1+C2+C3+…..+Cn
The resultant capacity of a number of capacitors, connected in parallel, is equal to the sum of their individual capacities.
(ii)V1= V2= V3 = V
(iii) q1 =C1V, q2 = C2V, q3 = C3V
(iv) Energy Stored, U = U1+U2+U3
(b)
(i) Capacitors in Series:- 1/C = 1/C1 + 1/ C2 +……+ 1/Cn
?The reciprocal of the resultant capacity of a number of capacitors, connected in series, is equal to the sum of the reciprocals of their individual capacities.
(ii) q1 = q2 = q3 = q
(iii) V1= q/C1, V2= q/C2, V3= q/C3
(iv) Energy Stored, U = U1+U2+U3
(a) Energy stored in a series combination of capacitors:-
W = ½ (q2/C1) + ½ (q2/C2) + ½ (q2/C3) = W1+W2+W3
Thus, net energy stored in the combination is equal to the sum of the energies stored in the component capacitors.
(b) Energy stored in a parallel combination of capacitors:-
W = ½ C1V 2 +½ C2V 2 + ½ C3V 2 = W1+W2+W3
The net energy stored in the combination is equal to sum of energies stored in the component capacitors.
(a) F = ½ ε0E2A
(b) F = σ2A/2ε0
(c) F=Q2/2ε0A
F = (Q2/2C2) (dC/dx) = ½ V2 (dC/dx)
V = [C1V1+ C2V2] / [C1+C2] = [Q1+Q2]/ [C1+C2]
ΔQ = [C1C2/C1+C2] [V1-V2]
ΔU = ½ [C1C2/C1+C2] [V1-V2] 2
(a) Q = Q0(1-e-t/RC)
(b) V = V0(1-e-t/RC)
(c) I = I0(1-e-t/RC)
(d) I0 = V0/R
(a) Q = Q0(e-t/RC)
(b) V = V0(e-t/RC)
(c) I = I0(e-t/RC)

Title:
Revision Notes on Electrostatic | askIITians
Meta description:
Revision Notes on electrostatic force, electrostatic field, electro static potential, electrostatic energy and capacitance provides by askIITians.
You Might Like to Refer:







