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Revision Notes on Electrostatic 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]
(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
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.
(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
(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
(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
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)
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