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(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)
(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
Electric potential energy U or work done of the system W of a three particle system having charge q1, q2 and q3:-
W = U = (1/4πε0) (q1q2/r12 + q1q3/r13 + q2q3/r23)
Electric potential energy of an electric dipole in an electric field:- Potential energy of an electric dipole, in an electrostatic field, is defined as the work done in rotating the dipole from zero energy position to the desired position in the electric field.
?
(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 ε0
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
(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)
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