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Physics 1

Capacitance

takriban dakika 4 kusoma

Mada za sehemu hiiElectrostaticsMada 3
  1. The electronic Field
  2. Electronic Potential
  3. Capacitance

Capacitors are circuit components used for storing charges. To store energy in a capacitor, electrons are transferred from one plate to the other so that one plate has a net negative charge and the other has an equal amount of positive charge. This process is called charging of a capacitor. The energy stored is given as:

Q=CVorC=QVQ = CV \quad \text{or} \quad C = \frac{Q}{V}

Capacitance of a Parallel Plate Capacitor

E=σε0=Qε0AandV=EdE = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \quad \text{and} \quad V = Ed

⇒C=ε0Ad\Rightarrow C = \frac{\varepsilon_0 A}{d}

If made of N plates:

C=(N−1)ε0AdC = \frac{(N - 1) \varepsilon_0 A}{d}

Factors Affecting Capacitance

  • Surface Area: C∝AC \propto A
  • Distance: C∝1dC \propto \frac{1}{d}
  • Dielectric Material: εr=CC0\varepsilon_r = \frac{C}{C_0}

Dielectric Influence

V=E(d−t)+EεrtandC=Aε0(d−t)+tεrV = E(d - t) + \frac{E}{\varepsilon_r} t \quad \text{and} \quad C = \frac{A\varepsilon_0}{(d - t) + \frac{t}{\varepsilon_r}}

If filled completely:

C=C0εrC = C_0 \varepsilon_r

Capacitors in Series

1Ceq=1C1+1C2+1C3+⋯+1Cn\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots + \frac{1}{C_n}

For two capacitors:

Ceq=C1C2C1+C2C_{eq} = \frac{C_1 C_2}{C_1 + C_2}

Capacitors in Parallel

Ceq=C1+C2+C3+⋯+CnC_{eq} = C_1 + C_2 + C_3 + \dots + C_n

Example Calculation

Given: Area = 100 cm² = 100×10−4 m2100 \times 10^{-4} \text{ m}^2, d = 1 mm = 1×10−3 m1 \times 10^{-3} \text{ m}, Q = 0.12μC0.12 \mu \text{C}, V = 120 V

C0=ε0Ad=8.854×10−12×100×10−41.0×10−3=8.854×10−11 FC_0 = \frac{\varepsilon_0 A}{d} = \frac{8.854 \times 10^{-12} \times 100 \times 10^{-4}}{1.0 \times 10^{-3}} = 8.854 \times 10^{-11} \text{ F}

C=QV=0.12×10−6120=1×10−9 FC = \frac{Q}{V} = \frac{0.12 \times 10^{-6}}{120} = 1 \times 10^{-9} \text{ F}

εr=CC0=1×10−98.854×10−11=11.3\varepsilon_r = \frac{C}{C_0} = \frac{1 \times 10^{-9}}{8.854 \times 10^{-11}} = 11.3

Series Capacitor Example

C1=6.0μF,C2=3.0μF,V=18 VC_1 = 6.0 \mu\text{F}, C_2 = 3.0 \mu\text{F}, V = 18 \text{ V}

a. Equivalent Capacitance:

Ceq=6×36+3=2μFC_{eq} = \frac{6 \times 3}{6 + 3} = 2 \mu\text{F}

b. Charge:

Q=CV=2×10−6×18=3.6×10−5 CQ = CV = 2 \times 10^{-6} \times 18 = 3.6 \times 10^{-5} \text{ C}

c. Voltages:

V1=QC1=3.6×10−56×10−6=6 V,V2=3.6×10−53×10−6=12 VV_1 = \frac{Q}{C_1} = \frac{3.6 \times 10^{-5}}{6 \times 10^{-6}} = 6 \text{ V}, \quad V_2 = \frac{3.6 \times 10^{-5}}{3 \times 10^{-6}} = 12 \text{ V}

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