The parallel plate capacitor is connected to the electrometer

When the electroscope metal ball is connected to one of the plates of the parallel plate capacitor, an equipotential body is formed; The metal shell of the electroscope and the other plate of the parallel plate capacitor are connected with the earth, and also constitute an equipotential body, then the electroscope becomes an electrometer, and the function of this electrometer is mainly to measure the voltage of the parallel plate capacitor, because the voltage between the metal ball of the electrometer and the metal shell (also called the potential difference) is the voltage between the two plates of the parallel plate capacitor (also called the potential difference).

However, at this time, the voltage of the capacitor can only be qualitatively judged by the angle of the pointer of the electrometer, and the specific value of the voltage is generally not known, and the capacitance c of the parallel plate capacitor is not known, so the amount of charge cannot be known by C=Q U.

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If the positive pole of the parallel-plate capacitor is connected to the electrostatic force, the positive electrode, the wire, and the metal sheet inside the electrostatic electrostatic are all equipotential and positively charged. The voltage between the sheet metal and its housing is equivalent to the voltage of the capacitor after it is grounded, and the higher the voltage, the more it will be deflected.

According to C=Q U, d increases, C decreases, and the amount of electricity in the capacitor is constant, so when U increases, the angle also increases.

The charge on a live parallel plate capacitor is transferred to the electrometer to deflect it.

The included angle is determined by the amount of charge, and the capacitance decreases when the distance is increased, and the amount of charge decreases when c=q u, so the included angle becomes smaller.

c. Analysis of test questions].

The electrometer is an instrument that uses electrostatic methods to quantitatively measure the electric charge and potential of the object, it is different from the electroscope, when the potential difference of the electrometer increases, the electric charge of the electrometer increases, and the angle between the pointers increases When the plate B moves up a little, it must be reduced by C = knowing C; From the meaning of the title, the capacitor charge Q does not change, then the potential difference between the two plates of the capacitor becomes larger from C=Q U, and the pointer angle of the electrometer increases at this time, so C. is selected

d.The positive area s decreases, and the capacitance c s decreases. There is no power exchange between the system and the outside world, the power does not change, and the voltage U=Q C increases. right？

Warm encouragement for questions from middle school students!

Analyze what happens when a battery is connected to a large parallel plate capacitor (assumed to be the ideal capacitor)? Naturally, the capacitor is electrified. Since the positive and negative charges are attracted to each other, when the battery is disconnected, these pairs of charges do not return to the battery and remain on the inner surface of the metal plates, maintaining equilibrium by the gravitational pull between the positive and negative charges.

This is the process of capacitor charging. This process is equivalent to the work done by an external electric field, which drives a part of the negative electrons on one plate (set a) that is originally balanced in pairs of positive and negative electrons to another plate (set as b) to make it a negative plate, and at the same time naturally the same amount of positive charge is on the plate a.

The answers to the questions are all in the analysis of the example above.

5. Why is the charge of the two adjacent plates not neutralized when the capacitor is connected in series? Answer: In which book did this sentence appear? This should be an unproblematic problem.

6. The left side of the leftmost plate and the right side of the rightmost plate are no longer electrified? Answer: Strictly speaking, the non-electrification mentioned here should mean that the docking site is not electrified. This does not mean that it is not charged to the other plates.

Because ideally the resistance of the voltmeter is infinite, the charge on B will not be lost due to the voltmeter grounding, and ideally, the charge on B will not be lost to the air.

So no matter how you move, the total amount of charge carried on b doesn't change.

But doing these actions affects the distribution of the charge on b. You can think of the voltmeter reading a value as one"Local voltage".And B has thickness, and if the charge on B gathers close to A, then the local voltage at the contact point of the voltmeter decreases.

Do you need to analyze ABCD one by one again?

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When the voltage between AB increases, then it is natural to know that the voltage between the ball above the electrometer and the shell increases, and the pointer naturally increases, and the pointer decreases.

A Reducing the distance increases the capacitance of AB, and when the amount of charge remains unchanged, then the voltage between AB naturally decreases, so the pointer decreases, correct.

b If you decrease the amount of charge on AB, then the voltage will naturally decrease, so it is also correct.

c The capacitance of capacitor AB is increased by inserting the dielectric, which is the same as A, so it is also correct.

D Move A up so that the area opposite A and B decreases, then the capacitance decreases, so the voltage should increase, and the pointer will also increase, so it is incorrect.

Answer: 1,A, correct, after disconnecting S, the capacitance C of a B separate capacitor decreases, U=Q C, so U increases, 2, B is wrong, S is closed, and U does not change.

3, C is wrong, S is closed, you are unchanged.

4, D error.

Choose A. When the switch is turned off, the total charge of the capacitor remains unchanged, the distance between abs increases, and the capacitance decreases according to the formula of the capacitor. q=cu, then u increases.

1.If the opening angle of the electrometer pointer increases, the potential difference between the plates increases, and the plate spacing increases, which leads to the capacitance of the capacitor decreasing with the increase of the board spacing.

2.If the opening angle of the electrometer pointer decreases, the potential difference between the plates decreases, and the board spacing decreases, which leads to the increase of the capacitance of the capacitor as the board spacing decreases.

3.In this process, the dielectric constant is increased.

The dielectric constant is reduced when the dielectric is withdrawn.

The opening angle is related to the amount of charge.

a. Increase the distance between the two plates, which is determined by the capacitance of the formula c=?s4πkd

It can be seen that the capacitance increases, the power of the capacitor remains unchanged, and it can be seen from the C=Qu analysis that the voltage between the plates increases, and the opening angle of the electrometer pointer becomes larger, so A is correct B. When the A board is moved up slightly, the positive area decreases, and the capacitance is determined by the formula C=?s4πkd

It can be seen that the capacitance decreases, and the power does not change, from c=q

u analysis can be seen that the voltage between the plates increases, and the opening angle of the electrometer pointer becomes larger, so b is correct c, the glass plate is inserted between the two plates, and the capacitance is determined by c = ?s4πkd

It can be seen that the capacitance increases, and the power does not change, from c=q

U analysis can be seen that the voltage between the plates decreases, the electrometer pointer opening angle becomes smaller, so C error D, if the A board is taken away, the charge on the B board remains unchanged, the electrometer is equivalent to an electroscope, and the pointer opening angle is not zero, so D is wrong

Therefore, ab is chosen

a. Parallel plate capacitors.

The middle is a uniform electric field, the field strength of A and B is the same, and the electric potential is different, and the electric potential of point A is higher, so A is wrong

b. The electric field strength of the metal inside the metal in the state of electrostatic equilibrium A and B.

are all zero, the whole conductor is an equipotential body, the electric potential is equal, so b is correct c, a, b are on the same equipotential surface, the electric potential is equal, and the direction of the electric field strength is different, so c is wrong d, according to the electric Zen object suspect field line.

The symmetry of the two points of AB shows that the electric potential of the two points of AB is the same, and the field strength is small, but the direction is opposite, so the electric field strength is different, so D is wrong

Therefore, b