Capacitor - Laboratory Review - General Physics | PHY 213, Lab Reports of Physics

Download Capacitor - Laboratory Review - General Physics | PHY 213 and more Lab Reports Physics in PDF only on Docsity! Phy213: General Physics III 5/20/2007 Chapter 25 Worksheet 1 Parallel Plate Capacitors: 1. Consider parallel plate capacitor (air filled) with a surface area of 225 cm2 and a charge of 1.5 µC (q) on each of its plates and a plate separation distance of 1.0x10-4 m. a. What is the capacitance of the capacitor? Ans. -9 o A C = = 2.0x10 F d ε b. What is the potential difference across the capacitor? Ans. q V = = 750V C c. How much energy is stored in the capacitor? Ans. 2 -41U = CV = 5.6x10 J 2 d. If the capacitor were filled with a dielectric material, κ=3.3 (while still maintaining the same amount of charge on the plates) what is the new capacitance? Ans. -9 airC = C = 6.6x10 Fκ e. How much charge would be stored in this capacitor with this dielectric material, at the same potential difference as in part (b)? Ans. -6q = CV = 5.0x10 C Cylindrical Capacitors: 2. A cylindrical air-filled capacitor (R1 = 0.5x10 -4m and R2 = 0.5x10 -2m) of length L=2.0 m has a potential difference of +120 V (outside – inside) between the inner and outer conductors. a. What is the capacitance of this capacitor? Ans. ( )2 1 -11o R R 2 L C = = 2.4x10 F ln πε b. How much charge is stored on each face of the capacitor? Ans. -9q = CV = 2.9x10 C c. Determine the equation for the electric field vector inside the capacitor. Ans. Apply Gauss’s Law for a Gaussian cylinder (noting that R1 & R2 << L): ˆE cylinder shaft r o o o q q = EA = E(2 rL) = E or i 2 rL 2 r λ π ε π ε π ε   Φ ⇒ ≅    
+ + + + + + + + + - - - - - - - - - Phy213: General Physics III 5/20/2007 Chapter 25 Worksheet 2 Spherical Capacitors: 3. A spherical air-filled capacitor (R1 = 0.5x10 -4m and R2 = 0.5x10 -2m) has a potential difference of +120 V (outside – inside) between the surfaces. a. What is the capacitance of this spherical capacitor? Ans. -151 2 o 2 1 R R C = 4 = 5.6x10 F R -R πε       b. How much charge is stored on each face of the capacitor? Ans. -13q = CV = 6.7x10 C c. Determine the equation for the electric field vector inside the capacitor. Ans. Apply Gauss’s Law for a “Gaussian sphere”: ˆ 2 E sphere r2 o o q q =EA =E(4 r )= E i 4 r π ε π ε Φ ⇒ ≅
4. Consider a spherical biological cell of radius 5.0 x 10-6 m. The membrane of the cell has capacitive properties no different that the electrolytic capacitors used in the lab. For this cell, the potential difference across the cell membrane (Vin-Vout) is -90 mV and the capacitance per unit surface area is 0.01 F/m2. a. What is the capacitance of the cell? Ans. ( )2 2 -12Fcell cellmC = 0.01 4 R = 3.1x10 Fπ b. How much charge is stored on each face of the membrane? Ans. -13 cellq = C V = 2.8x10 C c. How much energy is stored in the cell membrane? Ans. 2 -14 cell 1 U = C V = 1.3x10 J 2 d. If the dielectric constant for the cell membrane is κ = 9.0, what is the thickness of the cell membrane? Ans. since Rcell ≈ R1 ≈ R2, the cell capacitance can be approximated by: 2 2 -9cell cell o 2 1 o 2 1 R R C 4 or d = R -R 4 = 8.0x10 m R -R C πκε πκε     ≈ ≈       5. The measured capacitance for a blood cell is measured to be 1.6x10-2 pF. Assume that the membrane thickness and the dielectric constant is the same as for the previous problem. a. What is the surface area of the cell? Ans. 2 -12 2cell cell o o cell o 4 R A Cd C = A = = 1.6x10 m d d π κε κε κε     ≈ ⇒       b. Calculate the specific capacitance (capacitance per unit area] for this cell. Ans. 2 -14 F -12 2 m cell C 1.6x10 F = = = 0.010 A 1.6x10 m R R