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Rio Salado Community College *

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Electrical Engineering

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Feb 20, 2024

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Lesson 4 Problem Set 2 (V3) Name: Oluwadamilola Esan Capacitors PHY112 Section number 13534 32/40 Directions Show all work. Use appropriate labels and units. Highlight final answers (if numerical). 1. A parallel plate air capacitor has a plate separation distance of 5.6 mm , and the plate area measures 2.0 cm by 3.0 cm . What is the capacitance of the capacitor (3 points)? Change all values to meters: C= εA/d D = 0.0056m A= 0.02m*0.03m = 0.006m ε = 8.85 * 10 -12 F/m C = (8.85 * 10 -12 )(0.0006)/0.0056 C = 9.48 * 10 -13 Well done, though missing the unit, Farads, F, with the answer (-1) b) How much charge can this capacitor hold if connected to a 6 V battery (3 points)? Q = CV, C = 9.48 * 10 -13 , V = 6V Q = 9.48 * 10 -13 * 6.0 Q = 5.69 * 10 -12 C Correct 2. A parallel plate air capacitor has a capacitance of 2 F . The charge on each plate is 8 C . There are no batteries connected to the capacitor. How much work is required to double the separation distance between the plates? (6 pts) C= 2F Q= 8C V= Q / C V= 8C/2F V= 4V W= 1/2 Q(V2-V1 ) W= 1/2(8C)(2V-V) W= ½ (8C)(2(4V)-4V) W = ½ (8C)(4V)

W = 16 * 8C * 4V You were doing great up until this point. ½(8)(4)=16 (from line above), and that is the answer, 16 J (-1) W = 512C * V W = 512 C * (Q/C) W = 512Q W = 512 * 8C (substituted value) W= 4096 J 3. Two capacitors are connected in a series. They both have a plate separation distance of 0.015 m and a plate area of 0.0030 m 2 . They each have a different dielectric with dielectric constants of k 1 = 5.1 and k 2 = 9.2 , respectively. What is the overall capacitance of the system (6 points)? (Don’t be alarmed by how small the answers are. Capacitors in electronic circuits are typically very small). K 1 =5.1 K 2 = 9.2 E 0 = 8.854 * 10 -12 F/m A = 0.0030 m 2 d= 0.015m C 1 = k 1 E 0 A/d = 5.1 * ((8.854 * 10 -12 )*0.0030)/15 C 1 = 6.6 * 10 -9 F Correct numbers above, but the answer is 9.03E-12 F C 2 = k 2 E 0 A/d = 9.2 * ((8.854 * 10 -12 )*0.0030)/15 C 2 = 9.3 * 10 -9 F Again, correct numbers inserted, but it comes out to 1.63E-11 F 1/C = 1/C 1 + 1/C 2 1/C = 9.3 + 6.6/(6.6 * 10 -9 ) * (9.3 * 10 -9 ) This line doesn’t make sense. Just take the reciprocal of each of your values above, add those, and then take the reciprocal to get the final answer, which is 5.8E-12 F (-2) 1/C = 15.9/6.15*10 -17 C = 6.15*10 -17 /15.9 C = 3.87 * 10 -18 F b) If this system of capacitors is connected to a 9 V battery, how much total energy will be stored in the capacitors (4 points)? U = ½ CV 2 , C = 3.87 *10 -18 F, V=9V U = ½ * (3.87 * 10 -18 ) * 9 2

U = 1.57 * 10 -16 J Correct based on wrong answer in previous part 4. A parallel plate air capacitor with a capacitance of 3.0 nF is connected to a 6.0 V battery and charged. The capacitor is then disconnected from the battery and a dielectric with a dielectric constant of k = 8.0 is inserted between the plates. How much energy will be stored in the capacitor after inserting the dielectric (6 points)? U = ½ CV 2 V = 6V, C = 3.0 * 10 -9 F, k=8 U i = ½ * (3.0*10 -9 F) * (6.0V) 2 = 0.54 * 10 -9 J C’ = k * C U f = ½ * (8*3.0*10 -9 F) * (6.0V) 2 = 0.43 * 10 -6 J U f – U i = Energy = 0.43 * 10 -6 J - 0.54 * 10 -9 J Energy stored after is 0.43 * 10 -6 J As stated in section 4.2 of the lesson notes, the capacitance increases and the voltage decreases when a dielectric is inserted with the battery disconnected. The voltage is no longer 6.0 V. It is reduced by k. So you need to find the new capacitance and the new voltage first, and then use U=1/2CV 2 to find the energy. You could also find the charge, which doesn’t change, and then use U=1/2q 2 /C. The answer comes out to 6.75E-9 J (-2) 5. A dielectric is inserted between the plates of a charged parallel plate air capacitor with the battery disconnected. Describe whether each of the following will increase or decrease and explain your reasoning. a) Voltage difference between the plates (3 points) Voltage difference between the plates will decrease. The presence of a dielectric increases the capacitance, and according to the formula Q = CV, if C increases and Q remains constant, V must decrease. b) Electric field strength between the plates (3 points) Electric field strength between the plates will decrease. With a dielectric, the electric field is inversely proportional to the permittivity of the material (E= V/d). As the permittivity increases, the electric field decreases.

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c) Total charge stored on the plates (3 points) Total charge stored on the plates will remain constant. The capacitance (C=εA/d) increases with the dielectric, compensating for the decrease in voltage, resulting in constant charge (Q = CV). d) Energy stored in the capacitor (3 points) Energy stored in the capacitor will increase. The energy formula U = ½ CV 2 shows that with an increase in capacitance and a decrease in voltage, the energy stored in the capacitor will increase. Energy would decrease. U c =1/2qV 2 . The charge hasn’t changed, but the voltage has decreased. Thus, the stored energy must also decrease (-2)

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