A parallel plate capacitor is disconnected from the power supply after charging, and the negative plate is grounded After charging, a parallel plate capacitor is disconnected from the power supply, and the negative plate is grounded. A positive charge (small amount of charge) is fixed at point P between the two plates, as shown in the figure. E is the electric field strength between the two plates, u is the potential difference of the capacitor, and EP is the potential energy of the positive charge at point P. if the negative plate is kept stationary, move the positive plate to the position shown by the dotted line in the figure A. U becomes smaller, e remains unchanged, b.e becomes larger, EP becomes larger, c.u becomes smaller, EP remains unchanged, d.u remains unchanged, EP remains unchanged A I understand 1. What does this sentence mean when the negative plate is grounded 2. The explanation of C is that the electric potential does not change and the electric potential energy does not change Go to the position shown by the dotted line in the figure (move the positive plate downward)

A parallel plate capacitor is disconnected from the power supply after charging, and the negative plate is grounded After charging, a parallel plate capacitor is disconnected from the power supply, and the negative plate is grounded. A positive charge (small amount of charge) is fixed at point P between the two plates, as shown in the figure. E is the electric field strength between the two plates, u is the potential difference of the capacitor, and EP is the potential energy of the positive charge at point P. if the negative plate is kept stationary, move the positive plate to the position shown by the dotted line in the figure A. U becomes smaller, e remains unchanged, b.e becomes larger, EP becomes larger, c.u becomes smaller, EP remains unchanged, d.u remains unchanged, EP remains unchanged A I understand 1. What does this sentence mean when the negative plate is grounded 2. The explanation of C is that the electric potential does not change and the electric potential energy does not change Go to the position shown by the dotted line in the figure (move the positive plate downward)

When the negative electrode is grounded, the potential is zero. When it is fully charged, the electric quantity remains unchanged, the electric field strength between the plates remains unchanged, and the distance from P point to the negative plate remains unchanged, so the potential energy of the particle remains unchanged, the distance between the plates decreases, and the electric field strength remains unchanged, so the voltage between the two plates decreases

After charging, a parallel plate capacitor is disconnected from the power supply, and the negative plate is grounded. A positive charge between the two plates is fixed at point P, and the positive plate is on the top. When the positive plate is moved down, e will change

After charging, a parallel plate capacitor is disconnected from the power supply, and the positive plate is grounded. There is a positive charge between the two plates

A parallel plate capacitor is disconnected from the power supply after charging. If the distance between the two plates of the capacitor is increased by using an insulating handle
What is the change of electric field strength and why

After charging, it is disconnected from the power supply, and Q remains unchanged,
C=εS/4πkd,C=Q/U,E=U/d
Three way integration
When e = 4 π KQ / ε s, we can see that e is independent of the distance d between plates
Q does not change, e does not change

After charging, a parallel plate capacitor is disconnected from the power supply, and the positive plate is grounded. Between the two plates, there is a negative point charge (small charge) fixed at point P, as shown in the figure. E is the electric field strength between the two plates, u is the potential of the negative plate, and ε is the potential energy of the positive point charge at point P. move the positive plate to the position shown by the dotted line in the figure, then ()
A. E increases, u decreases, B. e remains unchanged, u increases, C. u increases, E decreases, D. u increases, e increases

A. When the parallel plate capacitor is disconnected from the power supply after charging, the electric quantity of the capacitor remains unchanged. When the positive plate is moved to the position shown by the dotted line in the figure, the distance between the plates decreases. According to the deduction formula e = 4 π KQ ɛ s, the electric field strength between the plates remains unchanged, the capacitance C increases, and the electric quantity Q remains unchanged. According to the formula u = QC, the voltage U between the plates decreases C. according to u = ed, e unchanged, the distance between P point and positive plate decreases, then the potential difference between positive plate and P point decreases, the potential of positive plate is zero, and the potential of P point is lower than that of positive plate, then the potential of P point increases, and the potential energy of positive charge at P point increases