Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Revision Notes on Electrostatic
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 = Kq_{1}q_{2}/r^{2}
Here, K = 1/4πε_{0} = 9×10^{9} Nm^{2}C^{-2} (in free space)
The relative permittivity (ε_{r}) of a medium is defined as the ratio between its permittivity of the medium (ε) and the permittivity (ε_{0}) of the free space.
ε_{r} = ε/ε_{0}
If q_{1}q_{2}>0, R.H.S is positive.
If q_{1}q_{2}<0, a negative sign from q_{1}q_{2} 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 = 3×10^{9} stat-coulomb
1 coulomb =(1/10) ab-coulomb (e.m.u of charge)
So, ε_{r} = ε/ε_{0} = F_{1}/F_{2}
Here, F_{1} and F_{2} 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
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/4πε_{0}) [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 = (1/4πε_{0}) [qx/(a^{2}+x^{2})^{3/2}]
Here σ is the surface charge.
(a) E_{out} = (1/4πε_{0}) (q/r^{2})
(b) E_{in} = 0
(a) Outside Point:- E_{out} = (1/4πε_{0}) (Q/r^{2})
(b) Inside Point:- E_{in} = (1/4πε_{0}) (Qr/R^{3})
(c) On the Surface:- E_{surface} = (1/4πε_{0}) (Q/R^{2})
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/2πε_{0}r
(b) Inside a point:- E = ρr/2ε_{0}
(a) Electric field at points outside the charged sheets:-
E_{P }= E_{R }= 0
(b) Electric field at point in between the charged sheets:-
E_{Q} = σ/ε_{0}
(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:-
= pE sinθ
Here, p is the dipole moment and θ is the angle between direction of dipole moment and electric field E.
E = λ/2πε_{0}r
The direction of electric field E is radially outward for a line of positive charge.
(a) Point at outside (r > R):- E = (1/4πε_{0}) (q/r^{2}), Here q is the total charge.
(b) Point at inside (r < R):- E = (1/4πε_{0}) (qr/R^{3}), Here q is the total charge.
E = σ/2ε_{0}
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/ε_{0}
P_{elec} = (½ε_{0}) σ^{2}
(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.
W_{AB} = q [V_{A}-V_{B}]
So, V = [V_{A}-V_{B}] = W/q
Units:- volt (S.I), stat-volt (C.G.S)
Dimension:- [V] = [ML^{2}T^{-3}A^{-1}]
Relation between volt and stat-volt:- 1 volt = (1/300) stat-volt
E = -dV/dx = --dV/dr
V = (1/4π ε_{0}) (q/r)
V = (1/4π ε_{0}) [q_{1}/r_{1} + q_{2}/r_{2} + q_{3}/r_{3}]
= V_{1}+V_{2}+ V_{2}+….
(a) Outside, V_{out} = (1/4π ε_{0}) (q/r)
(b) Inside, V_{in} = - (1/4π ε_{0}) (q/R)
(c) On the surface, V_{surface} = (1/4π ε_{0}) (q/R)
(b) Inside, V_{in} = (1/4π ε_{0}) [q(3R^{2}-r^{2})/2R^{3}]
(d) In center, V_{center} = (3/2) [(1/4π ε_{0}) (q/R)] = 3/2 [V_{surface}]
(a) Common potential, V = (1/4π ε_{0}) [(Q_{1}+Q_{2})/(r_{1}+r_{2})]
(b) q_{1} = r_{1}(Q_{1}+Q_{2})/(r_{1}+r_{2}) = r_{1}Q/ r_{1}+r_{2} ; q_{2} = r_{2}Q/ r_{1}+r_{2}
(c) q_{1}/q_{2} = r_{1}/r_{2} or σ_{1}/ σ_{2} = r_{1}/r_{2}
V (r,θ) = qa cosθ/4πε_{0}r^{2} = p cosθ/4πε_{0}r^{2}
(a) Point lying on the axial line:- V = p/4πε_{0}r^{2}
(b) Point situated on equatorial lines:- V = 0
(a) R = n^{1/3}r
(b) Q = nq
(c) V = n^{2/3}V_{small}
(d) σ = n^{1/3} σ_{small}
(e) E = n^{1/3} E_{small}
W = U = (1/4πε_{0}) (q_{1}q_{2}/r_{12}) = q_{1}V_{1}
W = U = (1/4πε_{0}) (q_{1}q_{2}/r_{12} + q_{1}q_{3}/r_{13} + q_{2}q_{3}/r_{23})
(a) If θ = 90º, then W = 0
(b) If θ = 0º, then W = -pE
(c) If θ = 180º, then W = pE
K. E = ½ mv^{2 }= 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^{-1}L^{-2}T^{4}A^{2}]
C = 4πε_{0}r
Capacitor:- A capacitor or a condenser is an arrangement which provides a larger capacity in a smaller space.
Capacity of a parallel plate capacitor:-
C_{air }= ε_{0}A/d
C_{med} = Kε_{0}A/d
Here, A is the common area of the two plates and d is the distance between the plates.
C = ε_{0}A/[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 = ε_{0}KA/ d
(a) C_{air} = 4πε_{0}R
(b) C_{med} = K (4πε_{0}R)
(a) When outer sphere is earthed:-
C_{air} = 4πε_{0} [ab/(b-a)]
C_{med} = 4πε_{0} [Kab/(b-a)]
(b) When the inner sphere is earthed:-
C_{1}= 4πε_{0} [ab/(b-a)]
C_{2} = 4πε_{0}b?
Net Capacity, C '=4πε_{0}[b^{2}/b-a]
Increase in capacity, ΔC = 4π ε_{0}b
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.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Solved Examples on Electrostatics:- Question...