Physics Answer Note #30

Question 1 A capacitor containing a dielectric with dielectric constant κ=4.0 has a capacitance of 12.0 nF . It is connected across a battery supplying a potential difference of 10.0 V . A) How much charge is stored on the capacitor? B) The capacitor is disconnected from the battery. Then the dielectric material is removed without discharging the capacitor. What is the potential difference between the plates now? Answer A capacitor is a passive electronic component that is used to store electrical energy in the form of an electric charge. It is made up of two conductive plates separated by a nonconductive material known as the dielectric. The capacitance of a capacitor is a measure of its ability to store charge and is influenced by several factors, such as the area of the plates, the distance between them, and the dielectric material used. In this scenario, the capacitor has a dielectric constant, κ, of 4.0 and a capacitance of 12.0 nF. It is also connected to a battery with a potential difference of 10.0 V. To answer the first question, we can use the formula Q = CV, where Q is the charge stored on the capacitor, C is the capacitance, and V is the potential difference. Substituting the given values, we get Q = (12.0 nF)(10.0 V) = 120 nC. Therefore, the capacitor has stored a charge of 120 nanoCoulombs. Moving on to the second question, when the capacitor is disconnected from the battery and the dielectric material is removed, the distance between the plates increases. This results in a decrease in the capacitance according to the formula C = κε₀A/d, where κ is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between them. As the dielectric material has been removed, the value of κ becomes 1, and assuming all other factors remain constant, the new capacitance is C' = (1)(8.85 x 10^-12 F/m)(A/d') = 8.85 x 10^-12 A/d'. Here, we are looking for the potential difference between the plates, which can be calculated using the formula V = Q/C'. Substituting the value of Q we found earlier, we get V = (120 nC)/(8.85 x 10^-12 A/d') = 13.56 V. In conclusion, when the capacitor is disconnected from the battery and the dielectric material is removed, the potential difference between the plates increases from 10.0 V to 13.56 V. This is because the removal of the dielectric material leads to an increase in the distance between the plates, resulting in a decrease in the capacitance and an increase in the potential difference. Question 2 A capacitor containing a dielectric with dielectric constant κ=4.0 has a capacitance of 12.0 nF . It is connected across a battery supplying a potential difference of 10.0 V . The capacitor is disconnected from the battery. Then the dielectric material is removed without discharging the capacitor. What is the new capacitance of the capacitor? Answer Dear fellow students, In order to accurately answer the given question, we must first understand the concept of capacitance and its relation to dielectric materials. A capacitor is an electrical component that stores energy in the form of an electric field between two conducting plates. The ability of a capacitor to store charge is measured by its capacitance, which is affected by both the physical properties of the capacitor and the material between the plates, known as the dielectric material. Now, let's focus on the specific scenario presented in the question. We are given a capacitor with a dielectric material of κ=4.0, a capacitance of 12.0 nF, and connected to a battery with a potential difference of 10.0 V. The first step in finding the new capacitance is to understand the effect of adding or removing a dielectric material on a capacitor's capacitance. When a dielectric material is added to a capacitor, it increases the electric field between the plates, thus increasing the capacitance. This is because the dielectric material, with its polar molecules, aligns with the electric field, allowing for more charge to be stored on the plates. On the other hand, removing the dielectric material decreases the electric field and subsequently decreases the capacitance. Keeping this in mind, let's tackle the question at hand. The capacitor is disconnected from the battery, meaning the charge stored on its plates remains constant. This also implies that the voltage between the plates remains constant as well. Additionally, the question specifies that the capacitor is removed without discharging it, indicating that the charge will remain the same. Therefore, in this scenario, we can use the formula C1V1 = C2V2, where C1 and V1 represent the initial capacitance and voltage, and C2 and V2 represent the final capacitance and voltage, respectively. By plugging in the given values, we get: (12.0 nF)(10.0 V) = C2(10.0 V) C2 = 12.0 nF/4.0 = 3.0 nF Hence, we can conclude that the new capacitance of the capacitor is 3.0 nF. To support this answer, let's refer to "Fundamentals of Physics" by Halliday and Resnick, where it is clearly explained that removing a dielectric material from a capacitor will decrease its capacitance. Similarly, the website "Hyperphysics" provides a detailed explanation of the effect of dielectric materials on capacitance and even offers a capacitance calculator for further calculation assistance. In conclusion, the new capacitance of the capacitor is 3.0 nF after the dielectric material is removed. This result can be achieved by using the formula C1V1 = C2V2, where the initial capacitance is 12.0 nF and the voltage remains constant at 10.0 V. "Fundamentals of Physics" by Halliday and Resnick and "Hyperphysics" are reliable sources that further support this answer and provide a better understanding of the concept. I hope this explanation was helpful in understanding the concept of capacitance and its relation to dielectric materials. Let's continue to strive for excellence in our academics. Best regards, [Your Name] Question 3 A capacitor containing a dielectric with dielectric constant κ=4.0 has a capacitance of 12.0 nF . It is connected across a battery supplying a potential difference of 10.0 V .A) How much charge is stored on the capacitor?B) The capacitor is disconnected from the battery. Then the dielectric material is removed without discharging the capacitor. What is the new capacitance of the capacitor? C) What is the potential difference between the plates now? Answer A) To find the amount of charge stored on the capacitor, we can use the formula Q = CV where Q is the charge, C is the capacitance and V is the potential difference. Plugging in the values given, we get: Q = (12.0 x 10^-9 F) x (10.0 V) = 1.2 x 10^-7 C Therefore, the charge stored on the capacitor is 1.2 x 10^-7 Coulombs. B) When the dielectric material is removed, the new capacitance of the capacitor can be calculated using the formula C = κε0A/d, where κ is the dielectric constant, ε0 is the permittivity of free space, A is the area of the plates and d is the distance between the plates. Since the dielectric constant is now 1 (since the dielectric material is removed), the new capacitance will be 4 times the original capacitance. Therefore, the new capacitance will be 12.0 nF x 4 = 48.0 nF. C) To find the potential difference between the plates now, we can use the formula V = Q/C, where Q is the charge and C is the capacitance. Since the charge remains constant and the new capacitance is 4 times the original, the potential difference will be 1/4th of the original potential difference. Therefore, the potential difference between the plates now will be 10.0 V / 4 = 2.5 V. Question 4 An electric dipole consists of two equal and opposite charges placed 2 𝑐𝑐 apart. When the dipole is placed in a uniform electric field of strength 10^5 𝑐/C, it experiences a maximum torque of 0.2 × 10^-3 𝑐𝑐 . Find the magnitude of each charge. Answer To find the magnitude of each charge in this scenario, we can use the formula for torque experienced by an electric dipole in an electric field: τ = pEsinθ, where τ is the torque, p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field vector. In this case, we are given the values for torque (0.2 × 10^-3 𝑐𝑐) and electric field strength (10^5 𝑐/C). We also know that the dipole moment of an electric dipole is equal to the product of the magnitude of one of the charges and the distance between the charges (p = qd). Therefore, we can rewrite the formula as: τ = qdEsinθ. Since the electric dipole is placed in a uniform electric field, the angle between the dipole moment and electric field is 90 degrees, making sinθ = 1. Plugging in the given values, we get 0.2 × 10^-3 𝑐𝑐 = q(2 𝑐𝑐)(10^5 𝑐/C)(1). Solving for q, we get the magnitude of each charge to be 1 × 10^-6 𝑐. Question 5 The network shown is assembled with uncharged capacitors 𝑐,𝑐, and Z , with 𝑐 𝑐 = 4 . 0 𝑐 𝑐 , 𝑐 𝑐 = 3 . 0 𝑐 𝑐 , and 𝑐 𝑐 = 5 . 0 𝑐 𝑐 . The switches 𝑐 1 and 𝑐 2 are initially open, and a potential difference 𝑐 𝑐 𝑐 = 1 2 0 𝑐 is applied between points a and 𝑐 . After the network is assembled, switch 𝑐 1 is then closed, but switch 𝑐 2 is kept open. What is the final potential difference across capacitor Z Answer The final potential difference across capacitor Z can be calculated using the equation 𝑐 = 𝑐/𝑐 , where 𝑐 is the potential difference, 𝑐 is the charge on the capacitor, and 𝑐 is the capacitance. Since capacitors X and Y are initially uncharged, the initial charge on capacitor Z is also 0. Therefore, the final potential difference across capacitor Z can be determined by finding the charge on capacitor Z after the switch S1 is closed. To find the charge on capacitor Z, we can use the formula 𝑐 = 𝑐𝑐 , where 𝑐 is the capacitance and 𝑐 is the potential difference applied. In this case, 𝑐 is equal to the potential difference between points a and b, which is 120V. Therefore, the charge on capacitor Z is given by 𝑐 = (5.0 𝑐𝑐)(120 V) = 600 𝑐C. Now, to calculate the final potential difference across capacitor Z, we can use the formula 𝑐 = 𝑐/𝑐. Plugging in the values, we get 𝑐 = (600 𝑐C)/(5.0 𝑐𝑐) = 120V. So, the final potential difference across capacitor Z is also equal to 120V. This is because the switches in the circuit are arranged in a series configuration, meaning that the capacitors are connected in a chain, and the voltage across each capacitor is equal to the potential difference applied (120V in this case). In summary, the final potential difference across capacitor Z in this circuit is 120V, and it remains unchanged even after switch S1 is closed. This is because the initial potential difference across capacitor Z is equal to the potential difference applied, and the arrangement of the switches in the circuit results in a series connection.