a) What is the charge on each electrode while the capacitor is attached to the b
ID: 2163829 • Letter: A
Question
a) What is the charge on each electrode while the capacitor is attached to the battery?Enter your answers numerically separated by a comma. Express your answer using two significant figures.
b)What is the electric field strength inside the capacitor while the capacitor is attached to the battery?
Express your answer using two significant figures.
c)What is the potential difference between the electrodes while the capacitor is attached to the battery?
Express your answer using two significant figures.
d)What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.9 apart? The electrodes remain connected to the battery during this process.
Enter your answers numerically separated by a comma. Express your answer using two significant figures.
e)What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.9 apart? The electrodes remain connected to the battery during this process.
Express your answer using two significant figures.
f)What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.9 apart? The electrodes remain connected to the battery during this process.
Express your answer using two significant figures.
Explanation / Answer
To solve this problem we must first find the capacitance (C) of this capacitor. It can be found by the formula C = k A / d, where A is the area of the plates (both have the same area) and d is the distance between them, and k is the dielectric constant for a vacuum (and air is close to it) which equals 8.85418 * 10**-12 farads per meter. So we obtain: C = 8.85418 times A divided by d to get the capacity in picofarads. We are given the distance d as 0.51 cm, or 0.0051 meter so we can find the area from the diameter by: A = ¼ p times the square of the diameter, or A = 0.785398 times (0.13)² which is A = 0.785398 * 0.0169 = 0.0132732 sq. meter. Then the capacitance C in picofarads equals: C = 8.85418 * 0.0132732 / 0.0051 = 23.043 picofarads, or 23.043 * 10**-12 fd. When the battery is disconnected, the charge Q on each electrode is found by Q = CV where V = 13 and C is 23.043 * 10**12, so Q = 2.9956 * 10**-10 coulomb. The electric field strength in volts per meter is 13 volts divided by 0.0051 meter, or 2549 volts per meter. The potential difference between the electrodes right after the battery is disconnected equals the battery voltage, which is 13 volts. If the electrodes are pulled apart using insulating handles so no electrons flow between them, the charge on each electrode is still 2.9956 * 10**-10 coulomb . But the capacity has changed; now it is 0.51/1.6, or 51/160 of its former value, so now it is 23.043 times 51/160, or 7.345 picofarads. Now the potential difference V is Q/C, or 40.784 volts. But the field strength is the same, since the charge is still the same. Now it is 40.784 volts divided by 0.016 meter, or 2549 volts per meter. Whew! Hope this answers all your questions.
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