When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates?
Four capacitors are connected as shown in Figure Find the equivalent capacitance between points a and b. Calculate the charge on each capacitor if ΔVab = 15.0 V.
Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00 μF.
A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled? . Therefore, the
Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm.
A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 × 108 V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4 000 V. Find the minimum area of the capacitor plates.
- A parallel plate capacitor made from 2 squares of metal, 2mm thick and 20cm on a side separated by 1mm with 1000V between them Find: a) capacitance b)charge per plate c) charge density d)electric field e) energy stored f) energy density
Consider the circuit as shown, where C1 = 6. 00mF and C2= 3 Consider the circuit as shown, where C1 = 6.00mF and C2= 3.00 mF and DV =20.0V. Capacitor C1 is first charged by closing of switch S1. Switch S1 is then opened and the charged capacitor is connected to the uncharged capacitor by the closing of S2. Calculate the initial charge acquired by C1 and the final charge on each. S1 close, S2 open C = Q/V Q = 120 mC After S1 open, S2 close Q1 + Q2 = 120 mC Same potential Q1 /C1 = Q2 / C2 (120-Q2)/C1= Q2/C2 (120 - Q2)/6 = Q2/ 3 Q2 = 40 mC Q 1= 80 mC
University Physics, Chapter 24 An isolated conducting sphere whose radius R is 6.85 cm has a charge of q=1.25 nC. How much potential energy is stored in the electric field of the charged conductor? Answer: Key Idea: An isolated sphere has a capacitance of C=4e0R The energy U stored in a capacitor depends on the charge and the capacitance according to … and substituting C=4pe0R gives April 16, 2017 University Physics, Chapter 24
An isolated conducting sphere whose radius R is 6 An isolated conducting sphere whose radius R is 6.85 cm has a charge of q = 1.25 nC. Question 2: What is the field energy density at the surface of the sphere? Answer: Key Idea: The energy density u depends on the magnitude of the electric field E according to so we must first find the E field at the surface of the sphere. Recall:
A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?
5- Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm.
University Physics, Chapter 24 An air-filled parallel plate capacitor has a capacitance of 1.3 pF. The separation of the plates is doubled, and wax is inserted between them. The new capacitance is 2.6pF. Question: Find the dielectric constant of the wax. Answer: Key Ideas: The original capacitance is given by Then the new capacitance is Thus rearrange the equation: April 16, 2017 University Physics, Chapter 24
University Physics, Chapter 24 Question 1: Consider a parallel plate capacitor with capacitance C = 2.00 F connected to a battery with voltage V = 12.0 V as shown. What is the charge stored in the capacitor? Question 2: Now insert a dielectric with dielectric constant = 2.5 between the plates of the capacitor. What is the charge on the capacitor? The additional charge is provided by the battery. April 16, 2017 University Physics, Chapter 24
Given a 7. 4 pF air-filled capacitor Given a 7.4 pF air-filled capacitor. You are asked to convert it to a capacitor that can store up to 7.4 J with a maximum voltage of 652 V. What dielectric constant should the material have that you insert to achieve these requirements?
One common kind of computer keyboard is based on the idea of capacitance. Each - key is mounted on one end of a plunger, the other end being attached to a movable metal plate. The movable plate and the fixed plate form a capacitor. When the key is pressed, the capacitance increases. The change in capacitance is detected, thereby recognizing the key which has been pressed. The separation between the plates is 5.00 mm, but is reduced to 0.150 mm when a key is pressed. The plate area is 9.50x10-5m2 and the capacitor is filled with a material whose dielectric constant is 3.50. Determine the change in capacitance detected by the computer.
University Physics, Chapter 24 If each capacitor has a capacitance of 5 nF, what is the capacitance of this system of capacitors? Find the equivalent capacitance We can see that C1 and C2 are in parallel, and that C3 is also in parallel with C1 and C2 We find C123 = C1 + C2 + C3 = 15 nF … and make a new drawing April 16, 2017 University Physics, Chapter 24
University Physics, Chapter 24 We can see that C4 and C123 are in series We find for the equivalent capacitance: … and make a new drawing = 3.75 nF April 16, 2017 University Physics, Chapter 24
University Physics, Chapter 24 We can see that C5 and C1234 are in parallel We find for the equivalent capacitance … and make a new drawing = 8.75 nF April 16, 2017 University Physics, Chapter 24
University Physics, Chapter 24 We have a parallel plate capacitor constructed of two parallel plates, each with area 625 cm2 separated by a distance of 1.00 mm. Question: What is the capacitance of this parallel plate capacitor? Answer: A parallel plate capacitor constructed out of square conducting plates 25 cm x 25 cm separated by 1 mm has a capacitance of about 0.5 nF. April 16, 2017 University Physics, Chapter 24
University Physics, Chapter 24 We have a parallel plate capacitor constructed of two parallel plates separated by a distance of 1.00 mm. Question: What area is required to produce a capacitance of 1.00 F? Then how is it possible to make large capacitance? Answer: Square conducting plates with dimensions 10.6 km x 10.6 km (6 miles x 6 miles) separated by 1 mm are required to produce a capacitor with a capacitance of 1 F. April 16, 2017 University Physics, Chapter 24
University Physics, Chapter 24 : A storage capacitor on a random access memory (RAM) chip has a capacitance of 55 nF. If the capacitor is charged to 5.3 V, how many excess electrons are on the negative plate? Answer: Idea: We can find the number of excess electrons on the negative plate if we know the total charge q on the plate. Then, the number of electrons n=q/e, where e is the electron charge in coulomb. Second idea: The charge q of the plate is related to the voltage V to which the capacitor is charged: q=CV. April 16, 2017 University Physics, Chapter 24
Capacitor C1 is connected across a battery of 5 V Capacitor C1 is connected across a battery of 5 V. An identical capacitor C2 is connected across a battery of 10 V. Which one has more charge? 1) C1 2) C2 3) both have the same charge 4) it depends on other factors Since Q = CV and the two capacitors are identical, the one that is connected to the greater voltage has more charge, which is C2 in this case.
1) increase the area of the plates 2) decrease separation between the plates 3) decrease the area of the plates 4) either (1) or (2) 5) either (2) or (3) What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? +Q –Q Since Q = CV, in order to increase the charge that a capacitor can hold at constant voltage, one has to increase its capacitance. Since the capacitance is given by , that can be done by either increasing A or decreasing d.
1) the voltage decreases 2) the voltage increases 3) the charge decreases 4) the charge increases 5) both voltage and charge change A parallel-plate capacitor initially has a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens? Since the battery stays connected, the voltage must remain constant! Since , when the spacing d is doubled, the capacitance C is halved. And since Q = CV, that means the charge must decrease. +Q –Q
1) 100 V 2) 200 V 3) 400 V 4) 800 V 5) 1600 V A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage? +Q –Q Once the battery is disconnected, Q has to remain constant, since no charge can flow either to or from the battery. Since , when the spacing d is doubled, the capacitance C is halved. And since Q = CV, that means the voltage must double.
What is the equivalent capacitance, Ceq , of the combination below? 1) Ceq = 3/2C 2) Ceq = 2/3C 3) Ceq = 3C 4) Ceq = 1/3C 5) Ceq = 1/2C What is the equivalent capacitance, Ceq , of the combination below? The 2 equal capacitors in series add up as inverses, giving 1/2C. These are parallel to the first one, which add up directly. Thus, the total equivalent capacitance is 3/2C. o C Ceq
1) V1 = V2 2) V1 > V2 3) V1 < V2 4) all voltages are zero How does the voltage V1 across the first capacitor (C1) compare to the voltage V2 across the second capacitor (C2)? The voltage across C1 is 10 V. The combined capacitors C2 + C3 are parallel to C1. The voltage across C2 + C3 is also 10 V. Since C2 and C3 are in series, their voltages add. Thus the voltage across C2 and C3 each has to be 5 V, which is less than V1. C1 = 1.0 mF C3 = 1.0 mF C2 = 1.0 mF 10 V
1) Q1 = Q2 2) Q1 > Q2 3) Q1 < Q2 4) all charges are zero How does the charge Q1 on the first capacitor (C1) compare to the charge Q2 on the second capacitor (C2)? We already know that the voltage across C1 is 10 V and the voltage across both C2 and C3 is 5 V each. Since Q = CV and C is the same for all the capacitors, we have V1 > V2 and therefore Q1 > Q2. C1 = 1.0 mF C3 = 1.0 mF C2 = 1.0 mF 10 V