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Test: Capacitors - MCAT MCQ


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10 Questions MCQ Test - Test: Capacitors

Test: Capacitors for MCAT 2024 is part of MCAT preparation. The Test: Capacitors questions and answers have been prepared according to the MCAT exam syllabus.The Test: Capacitors MCQs are made for MCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Capacitors below.
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Test: Capacitors - Question 1
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In a parallel-plate capacitor, how can the capacitance be decreased?

Detailed Solution for Test: Capacitors - Question 1
  • Recall the capacitance formula for a parallel plate capacitor.
  • The formula is C = ε0A/d, with A being the surface area of one of the plates and d being the distance between two plates. Note that capacitance can be inferred purely from geometric properties of the capacitor.
  • Increasing the gap between in charged plates will decrease the capacitance.
Test: Capacitors - Question 2
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Imagine a parallel-plate capacitor with a plate separation of d = 1 cm. The plates are both right triangles with both base and height equal to 2 cm. What is the capacitance of this capacitor? (assume 

Detailed Solution for Test: Capacitors - Question 2
  • Recall the capacitance formula for a parallel plate capacitor:
  • We need to find surface area A of each plate,



  • The capacitance of this capacitor is 2 × 10−13F
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Test: Capacitors - Question 3
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A parallel plate capacitor separated 10 cm, by an air barrier is connected to a 100 V battery. The capacitance of the capacitor is 1 picofarad while the battery is connected. Without disconnecting the battery, the parallel plates are moved so they are now 20 cm apart. What happens to the energy stored in the capacitor?

Detailed Solution for Test: Capacitors - Question 3
  • Since this question does not ask for actual numeric answers, it's easier to determine relationships rather than "plugging-and-chugging" for the actual answer. Recall the formula for energy stored in a capacitor: U = 0.5QV. Try to express this in terms of C.
  • Given Q = CV, U = 0.5 CV2. You know that  so substitute that in.
  • Since the potential difference is maintained, you should see that  d increases from 10 cm to 20 cm. 
  • The energy stored in the capacitor decreases by half when the plates are moved from 10 cm to 20 cm apart.
Test: Capacitors - Question 4
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What is the equivalent capacitance of this circuit? 
Detailed Solution for Test: Capacitors - Question 4
  • Recall equivalent capacitance in a circuit where the elements are connected in parallel.
  • Cequivalent = C1 + C2 + C3.
  • The equivalent capacitance is 6 pF.
Test: Capacitors - Question 5
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What is the equivalent capacitance of this circuit?
Detailed Solution for Test: Capacitors - Question 5
  • Recall equivalent capacitance in a circuit where the elements are connected in series.
  • To find Cequivalent, you must take the reciprocal of both sides. The equivalent capacitance is 2/3 pF.
Test: Capacitors - Question 6
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What is the correct order of the capacitances of these circuits, from lowest to highest? Assume all capacitors have the same capacitance.
Detailed Solution for Test: Capacitors - Question 6
  • For convenience, we’ll say that the capacitance of each capacitor is 1 F; the answer is independent of what value we pick.
  • We instantly can infer the capacitance of choice “C” using the series addition formula, 1/Ctotal = 1 + 1 and so Ctotal = 1/2.
  • For circuit “D,” we first use the series formula for the bottom rung, 1/Cbottom = 1/1 + 1/1 F, to find that Cbottom = 1/2 F. We then use the parallel formula to add this rung to the capacitance of the top rung for a total capacitance of Ctotal  = 1/2 + 1 = 3/2 F
  • We can determine the capacitance of circuit “A” by first combining the two capacitors in parallel Ctotal  = 1 + 1 = 2 F. We then add this in series with the capacitor on the left to find 1/Ctotal = 1/2 + 1 F, and so Ctotal = 2/3 F
  • We recognize that circuit “B” is the same as circuit “D”, except we have to add another capacitor in series with the entirety of choice D. 1/Ctotal = 2/3 + 1 and so Ctotal = 3/5 F
  • Combining our results, we find C < B < A < D
Test: Capacitors - Question 7
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If you are given a large parallel-plate capacitor containing just air as a dielectric, which of the following tools would be sufficient to determine the capacitor’s capacitance?
Detailed Solution for Test: Capacitors - Question 7
  • Think about which formulas you know for capacitors: the fact that the question specifies parallel-plate means that you can use the formula C = ε0 A/d in addition to Q = CV.
  • It is impossible to infer the capacitance purely from the voltage or resistance of a circuit with a capacitor; you must also know the charge on the capacitor.
  • The plate area and separation are both purely geometric properties, and so a ruler is all you need to determine the capacitance.
Test: Capacitors - Question 8
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Which of the following parallel plate capacitors will store the most charge on the positive plate when a potential difference V is applied across the plate?
Detailed Solution for Test: Capacitors - Question 8
  • A dielectric is a material that is placed between parallel plate capacitors to increase the capacitance of the capacitor. Glass and vacuum are examples of dielectric material.
  • Capacitance increases by the dielectric constant as: Cnew = C*K. Vacuum is the most basic dielectric at κ = 1. Almost everything you encounter on the test that is not air or vacuum will have a higher dielectric constant (glass is 5). Given C = ε0A/d and Q = CV, we can have the modified Q = Kε0AV/d  for a capacitor with a dielectric.
  • The parallel plate capacitor with plates separated by glass of thickness L/2 will store the most charge.
Test: Capacitors - Question 9
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Four capacitors of 10 pF are connected in parallel with two of the capacitors having a Mylar dielectric (κ = 3) of thickness L inserted and two capacitors only separated by air at a distance L. A voltage potential of 40 V is applied across the circuit. How does the charge stored on the capacitors differ between the capacitors with the Mylar and without?
Detailed Solution for Test: Capacitors - Question 9
  • Dielectrics help increase the capacitance of the capacitor, allowing additional charge to be stored on the capacitor.
  • Capacitance increases by the dielectric constant as: Cnew = C*K. Given C = ε0A/d and Q = CV, we can have the modified Q = Kε0 AV/d for a capacitor with a dielectric.
  • The Mylar capacitors store 3 times as much charge as the air capacitors.
Test: Capacitors - Question 10
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Four capacitors of 10 pF are connected in parallel with two of the capacitors having a Mylar dielectric (κ = 3) of thickness L inserted and two capacitors only separated by air at a distance L. A voltage potential of 40 V is applied across the circuit and subsequently disconnected. The Mylar dielectrics are then removed and charges are allowed to equilibrate in the system. What is the energy stored on each capacitor?
Detailed Solution for Test: Capacitors - Question 10
  • When four capacitors are connected in parallel, they all experience the same voltage 40V. We are interested in U, which is given by 0.5*Q2/C.
  • So what is Q? For the capacitors without the Mylar, it would be Q = CV or 400 pFV (or pC, since a coulomb = FV). For the capacitors with the Mylar, it would be Q = 3*CV or 1200 pC.
  • When the voltage potential and the Mylar dielectrics are removed, the charge in the circuit will redistribute equally to all of the different capacitors. Initially, we have 2·4 + 2·1200, so a total of 3200 pC in charge distributed among 4 capacitors. When charges are allowed to freely equilibrate, we have 800 pC on each capacitor. We can use this to find the energy stored on each capacitor.
  • Use U = 0.5*Q2 /C and we have 32000 pJ (a joule is equivalent to coulomb squared divided by farad).
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