Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Cylindrical Spherical Capacitors Series and Parallel Combination
→ \rightarrow → → \rightarrow → Cylindrical Spherical Capacitors Series and Parallel Combination → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Parallel Plate Capacitor
Consists of two plates of area A separated by distance d.
Between these two plates there is an electric field.
ϵ 0 \epsilon_0 ϵ 0
C = ϵ 0 A d \frac{\epsilon_0 A}{d} d ϵ 0 A
→ \rightarrow → → \rightarrow → Parallel Plate Capacitor → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Cylindrical Capacitor
λ \lambda λ
There is no Gaussian flux, electric flux on the top and bottom surface of the cylinder of the Gaussian surface, there is only flux from the cylindrical surface
2 π r ⋅ l ⋅ E = λ l ϵ 0 2 \pi r \cdot l \cdot E =\frac{\lambda l}{\epsilon_0} 2 π r ⋅ l ⋅ E = ϵ 0 λ l
E ⃗ = λ 2 π ϵ 0 r r ^ \vec{E} =\frac{\lambda}{2 \pi \epsilon_0 r} \hat{r} E = 2 π ϵ 0 r λ r ^
Cylindrical Spherical Capacitors Series and Parallel Combination → \rightarrow → → \rightarrow → Cylindrical Capacitor → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Potential Difference
V = V ( a ) − V ( b ) V =V(a)-V(b) V = V ( a ) − V ( b )
= − ∫ b a E ⃗ ⋅ d r ⃗ =-\int_b^a \vec{E} \cdot d \vec{r} = − ∫ b a E ⋅ d r = λ 2 π ϵ 0 ∫ a b d r r =\frac{\lambda}{2 \pi \epsilon_0} \int_a^b \frac{d r}{r} = 2 π ϵ 0 λ ∫ a b r d r
V = λ 2 π ϵ 0 ln ( b a ) V=\frac{\lambda}{2 \pi \epsilon_0} \ln \left(\frac{b}{a}\right) V = 2 π ϵ 0 λ ln ( a b )
Length L : charge Q=λ \lambda λ
Parallel Plate Capacitor → \rightarrow → → \rightarrow → Potential Difference → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitance
V = Q 2 π ϵ 0 L ln ( b a ) V=\frac{Q}{2 \pi \epsilon_0 L} \ln \left(\frac{b}{a}\right) V = 2 π ϵ 0 L Q ln ( a b )
V = Q C V=\frac{Q}{C} V = C Q
Capacitance C = 2 π ϵ 0 L ln ( b / a ) \frac{2 \pi \epsilon_0 L}{\ln (b / a)} l n ( b / a ) 2 π ϵ 0 L
Capacitance per unit length = C L = 2 π ϵ 0 ln ( b / a ) \frac{C}{L}=\frac{2 \pi \epsilon_0}{\ln (b / a)} L C = l n ( b / a ) 2 π ϵ 0
Cylindrical Capacitor → \rightarrow → → \rightarrow → Capacitance → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Coaxial Cables
Example
≃ 80 p F / m = 80 × 10 − 12 F / m \simeq 80 \mathrm{pF} / \mathrm{m}=80 \times 10^{-12} \mathrm{F} / \mathrm{m} ≃ 80 pF / m = 80 × 1 0 − 12 F / m
Coaxial cables C ∼ 70 C \sim 70 C ∼ 70
Potential Difference → \rightarrow → → \rightarrow → Coaxial Cables → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Spherical Capacitor
The inner sphere is charged positively, assume to be charged positively the outer sphere is assumed to be charged negatively
Radius of inner sphere = r a \ r_a r a
Radius of outer sphere = r b \ r_b r b
Electric Flux = 4 π r 2 4 \pi r^2 4 π r 2
Capacitance → \rightarrow → → \rightarrow → Spherical Capacitor → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Potential Difference
V = − ∫ r b r a E ⃗ ⋅ d r ⃗ V =-\int_{r_b}^{r_a} \vec{E} \cdot d \vec{r} V = − ∫ r b r a E ⋅ d r = ∫ r a r b d 4 π ϵ o r 2 d r =\int_{r_a}^{r_b} \frac{d}{4 \pi \epsilon_o r^2} d r = ∫ r a r b 4 π ϵ o r 2 d d r
= Q 4 π ϵ 0 ( − 1 r ) r a r b =\frac{Q}{4 \pi \epsilon_0}\left(-\frac{1}{r}\right)_{r_a}^{r_b} = 4 π ϵ 0 Q ( − r 1 ) r a r b
= Q 4 π ϵ 0 ( 1 r a − 1 r b ) =\frac{Q}{4 \pi \epsilon_0}\left(\frac{1}{r_a}-\frac{1}{r_b}\right) = 4 π ϵ 0 Q ( r a 1 − r b 1 )
V = Q 4 π ϵ 0 ( r b − r a ) r a r b V =\frac{Q}{4 \pi \epsilon_0} \frac{\left(r_b-r_a\right)}{r_a r_b} V = 4 π ϵ 0 Q r a r b ( r b − r a )
Coaxial Cables → \rightarrow → → \rightarrow → Potential Difference → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitance
Q is the charge carried by the conductors
C = Q V \frac{Q}{V} V Q 4 π ϵ 0 ( r a r b ) ( r b − r a ) \frac{4 \pi \epsilon_0\left(r_a r_b\right)}{\left(r_b-r_a\right)} ( r b − r a ) 4 π ϵ 0 ( r a r b )
Both conductors carry the same amount of charge Q
Spherical Capacitor → \rightarrow → → \rightarrow → Capacitance → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitance
Capacitance of a single sphere conducting sphere of radius r a.
The other conductor which is supposed to be carrying negative charges as infinite size
Limit r b → ∞ r_b \rightarrow \infty r b → ∞
C = 4 π ϵ o r a 4 \pi \epsilon_o r_a 4 π ϵ o r a
Potential Difference → \rightarrow → → \rightarrow → Capacitance → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Spherical Capacitor
r a r_a r a r b = ∞ r_b = \infty r b = ∞
C= 4 π ϵ 0 r a 4 \pi \epsilon_0 r_a 4 π ϵ 0 r a
= 10 − 3 9 × 10 9 ≃ 0.11 p F = 0.11 × 10 − 12 F \frac{10^{-3}}{9 \times 10^9} \simeq 0.11 \mathrm{pF}=0.11 \times 10^{-12} \mathrm{F} 9 × 1 0 9 1 0 − 3 ≃ 0.11 pF = 0.11 × 1 0 − 12 F
If I put a charge of 1 pC then the voltage generated.
V = Q C = 10 − 12 0.11 × 10 − 12 ≈ 9 V V=\frac{Q}{C}=\frac{10^{-12}}{0.11 \times 10^{-12}} \approx 9 \mathrm{V} V = C Q = 0.11 × 1 0 − 12 1 0 − 12 ≈ 9 V
Capacitance → \rightarrow → → \rightarrow → Spherical Capacitor → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitance of The Earth
Assume the earth to be spherical
R = 6371 km
C = 4 π ϵ 0 ⋅ R C = 4 \pi \epsilon_0 \cdot R C = 4 π ϵ 0 ⋅ R = 6.371 × 10 6 9 × 10 9 =\frac{6.371 \times 10^6}{9 \times 10^9} = 9 × 1 0 9 6.371 × 1 0 6
≃ 7.08 × 10 − 4 F \simeq 7.08 \times 10^{-4} \mathrm{F} ≃ 7.08 × 1 0 − 4 F = 708 μ F =708 \mu \mathrm{F} = 708 μ F
Capacitance → \rightarrow → → \rightarrow → Capacitance of The Earth → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitors Connected in Series
They are not connected to any other part of the circuit. So, the net charge within this must be equal to 0
V = V 1 + V 2 + V 3 V=V_1+V_2+V_3 V = V 1 + V 2 + V 3
Charge supplied = q
V 1 = q C 1 ; V 2 = q 1 C 2 ; V 3 = q C 3 V_1=\frac{q}{C_1} ; \quad V_2=\frac{q_1}{C_2} ; \quad V_3=\frac{q}{C_3} V 1 = C 1 q ; V 2 = C 2 q 1 ; V 3 = C 3 q
Spherical Capacitor → \rightarrow → → \rightarrow → Capacitors Connected in Series → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitors Connected in Series
V = V 1 + V 2 + V 3 V=V_1+V_2+V_3 V = V 1 + V 2 + V 3 = q C 1 + q C 2 + q C 3 = \frac{q}{C_1} + \frac{q}{C_2} + \frac{q}{C_3} = C 1 q + C 2 q + C 3 q
C = q V C = \frac{q}{V} C = V q
V q = 1 C 1 + 1 C 2 + 1 C 3 \frac{V}{q}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} q V = C 1 1 + C 2 1 + C 3 1
1 C = 1 C 1 + 1 C 2 + 1 C 3 \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} C 1 = C 1 1 + C 2 1 + C 3 1
n Capacitor connected in series.
1 C = ∑ i = 1 N 1 C i \frac{1}{C} =\sum_{i=1}^{N} \frac{1}{C_i} C 1 = ∑ i = 1 N C i 1 = 1 C 1 + 1 C 2 + ⋯ + 1 C n =\frac{1}{C_1}+\frac{1}{C_2}+\cdots+\frac{1}{C_n} = C 1 1 + C 2 1 + ⋯ + C n 1
Capacitance of The Earth → \rightarrow → → \rightarrow → Capacitors Connected in Series → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitances in Parallel
Capacitor have a potential difference of V
Charge on C 1 = q 1 C_1 = q_1 C 1 = q 1
Charge on C 2 = q 2 C_2 = q_2 C 2 = q 2
Charge on C 3 = q 3 C_3 = q_3 C 3 = q 3
Capacitors Connected in Series → \rightarrow → → \rightarrow → Capacitances in Parallel → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitances in Parallel
Total Charge Supplied
q = q 1 + q 2 + q 3 q = q_1 + q_2 + q_3 q = q 1 + q 2 + q 3
q 1 = C 1 V , q 2 = C 2 V , q 3 = q 3 V q_1 = C_1V , q_2 = C_2V , q_3 = q_3V q 1 = C 1 V , q 2 = C 2 V , q 3 = q 3 V
q = ( C 1 + C 2 + C 3 ) V q =(C_1+C_2+C_3)V q = ( C 1 + C 2 + C 3 ) V
C = q V C = \frac{q}{V} C = V q
Capacitors Connected in Series → \rightarrow → → \rightarrow → Capacitances in Parallel → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitances in Parallel
C = C 1 + C 2 + C 3 C = C_1 + C_2 + C_3 C = C 1 + C 2 + C 3
For n capacitor in parallel.
C = ∑ i = 1 n C i C=\sum_{i=1}^n C_i C = ∑ i = 1 n C i
Capacitances in Parallel → \rightarrow → → \rightarrow → Capacitances in Parallel → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Capacitances in Parallel
Capacitances in Parallel → \rightarrow → → \rightarrow → Capacitances in Parallel → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Problem
Example
C 23 = C 2 + C 3 C_\text{23} = C_2 +C_3 C 23 = C 2 + C 3
C ⇒ C \Rightarrow C ⇒
1 C = 1 C 1 + 1 C 2 + C 3 \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2+C_3} C 1 = C 1 1 + C 2 + C 3 1
C 1 = 25 μ f , C 2 = 5 μ f , C_1=25 \mu \mathrm{f}, \quad C_2=5 \mu \mathrm{f}, C 1 = 25 μ f , C 2 = 5 μ f ,
C 23 = C 2 + C 3 = 25 μ F C_{23}=C_2+C_3=25 \mu \mathrm{F} C 23 = C 2 + C 3 = 25 μ F
Capacitances in Parallel → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Problem
C 23 = C 2 + C 3 C_\text{23} = C_2 +C_3 C 23 = C 2 + C 3
1 C = 1 C 1 + 1 C 23 \frac{1}{C} =\frac{1}{C_1}+\frac{1}{C_{23}} C 1 = C 1 1 + C 23 1 = 1 25 + 1 25 =\frac{1}{25}+\frac{1}{25} = 25 1 + 25 1 = 2 25 =\frac{2}{25} = 25 2
C = 25 2 = 12.5 μ F C =\frac{25}{2}=12.5 \mu \mathrm{F} C = 2 25 = 12.5 μ F
Capacitances in Parallel → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Problem
V = 10 V V = 10V V = 10 V
q = C V = 12.5 × 10 − 6 × 10 V = 125 μ c q=CV=12.5 \times 10^{-6} \times 10 \mathrm{V}=125 \mu \mathrm{c} q = C V = 12.5 × 1 0 − 6 × 10 V = 125 μ c
V 1 = q C 1 = 125 × 10 − 6 25 × 10 − 6 = 5 V V_1=\frac{q}{C_1}=\frac{125 \times 10^{-6}}{25 \times 10^{-6}}=5 \mathrm{V} V 1 = C 1 q = 25 × 1 0 − 6 125 × 1 0 − 6 = 5 V
q 2 = C 2 V = 5 × 10 − 6 × 5 = 25 μ C q_2=C_2 V=5 \times 10^{-6} \times 5=25 \mu C q 2 = C 2 V = 5 × 1 0 − 6 × 5 = 25 μ C
q 3 = C 3 V = 20 × 10 − 6 × 5 = 100 μ C q_3=C_3 V=20 \times 10^{-6} \times 5=100 \mu C q 3 = C 3 V = 20 × 1 0 − 6 × 5 = 100 μ C
We can use the law for adding capacitors connected in series or in parallel and find the equivalent capacitor.
Problem → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Problem
Consider a parallel plate capacitor as shown:
You have two conducting plates separated by distance d.
A solid metallic slab of thickness d 2 \frac{d}{2} 2 d
Problem → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Thank You
Problem → \rightarrow → → \rightarrow → Thankyou → \rightarrow → → \rightarrow →
Resume presentation
Cylindrical-Spherical-Capacitors-Series-Parallel-Combination L-1 Cylindrical Spherical Capacitors Series and Parallel Combination $\rightarrow$ $\rightarrow$ Cylindrical Spherical Capacitors Series and Parallel Combination $\rightarrow$ Parallel Plate Capacitor $\rightarrow$ Cylindrical Capacitor
