As shown in the figure, D is an ideal diode (the forward resistance is 0, the reverse resistance is infinite), and the parallel plate capacitor is connected to the power supply with constant electromotive force through the diode. The following is the statement about the electric quantity Q on the plates and the electric field strength e between the plates changing with the distance between the plates: ① the distance between the plates becomes smaller, Q increases; ② the distance between the plates becomes smaller, e increases; ③ the distance between the plates becomes larger, Q decreases; ④ When the distance between plates increases, e does not change A. ①②B. ①②③C. ①②④D. ①②③④

As shown in the figure, D is an ideal diode (the forward resistance is 0, the reverse resistance is infinite), and the parallel plate capacitor is connected to the power supply with constant electromotive force through the diode. The following is the statement about the electric quantity Q on the plates and the electric field strength e between the plates changing with the distance between the plates: ① the distance between the plates becomes smaller, Q increases; ② the distance between the plates becomes smaller, e increases; ③ the distance between the plates becomes larger, Q decreases; ④ When the distance between plates increases, e does not change A. ①②B. ①②③C. ①②④D. ①②③④

① When the distance between plates becomes smaller, the capacitance increases and the voltage remains unchanged. From q = Cu, the capacitor charges when the electric quantity increases. Therefore, ① correct; ③ As the distance between plates increases, the capacitance decreases, and the voltage remains unchanged. Because of the unidirectional conductivity of the diode, the capacitor cannot discharge, and the electric quantity Q remains unchanged. Therefore, error (3). & nbsp; & nbsp; & nbsp; & nbsp; (4) as the distance between plates increases, the capacitance decreases, and the capacitor cannot discharge, and the electric quantity Q remains unchanged. According to the inference e = 4 π KQ ɛ s, the electric field strength e between plates remains unchanged. Therefore, error (4) is correct, so C is selected

In the circuit as shown in the figure, the electromotive force of the power supply is e, the internal resistance is ignored, R1, R2, R3 and R4 are constant resistance, C is capacitor, and switch S is open,
If the switch S is closed, what is the charge passing through R4 in a long time after s is closed?

S open: UR2 = E / R1 + R2 C = Q1 / u Q1 = Cu = (CE / R1 + R2) * r2s close: ur1 = (E / R1 + R2) R1 C = Q2 / u Q2 = Cu = (CE / R1 + R2) * r1q '= Q1 + Q2 = (CE / R1 + R2) * R1 + (CE / R1 + R2) * R2 = (CE / R1 + R2) * (R1 + R2) = CE

As shown in the figure, the electromotive force of the power supply is e = 10V, the internal resistance is ignored, and the four resistors R1 = R2 = R3 = R4 = 1 ohm. The capacitance of the capacitor is C = 1uF
The switch S is first opened. If s is closed, what is the direction of the current flowing through the ammeter?
This is a picture

Before the switch is closed, the power supply charges the capacitor so that the voltage at both ends of the capacitor is equal to the power supply voltage. If the negative charge accumulates from the negative pole of the power supply to the lower plate of the capacitor, the current is from bottom to top. If the voltage at both ends of the capacitor is equal to the R2 voltage after the switch is closed, the capacitor will discharge to the outside, so the current is reverse, from top to bottom

As shown in the figure, the electromotive force E of the power supply is 12V, the internal resistance is ignored, the resistance R1 = R2 = R3 = 6 Ω, R4 = 12 Ω, the capacitance C of the capacitor is 10 μ F, and a charged particle in the capacitor is just in a static state. If the resistance R2 suddenly breaks in the process of operation, the ammeter can be regarded as an ideal ammeter
A. Before R2 breaking, UAB = 2VB. In the process of change, the direction of current flowing through the ammeter is from bottom to top. C. After R2 breaking, the acceleration of static charged particles accelerating downward is 3gd. In the process of resistance R2 breaking to circuit stability, the amount of electricity flowing through the ammeter is 8 × 10-5c

A. Before the break of R2, U1 = U3 = 6V, U2 = 4V, UAB = φ a - φ B = - 2V, so a is wrong; before the break of B and R2, UAB = - 2V, the upper plate of capacitor is negatively charged, and the lower plate is positively charged; after the break of R2, U1 ′ = U3 ′ = 6V, UAB ′ = U3 ′ = 6V, the upper plate of capacitor is positively charged, and the lower plate is negatively charged