I was a beginner in electricity and wanted to ask some questions about physics and electricity in my

The charge is evenly distributed on the outer surface, and the electric field in the cavity is 0, which can be distributed in any case if it is not a conductor.

du=e*x, the distance between the equipotential surfaces, multiplied by the electric field strength is equal to the electric potential difference, in other words, the electric potential difference is constant, and the stronger the electric field, the smaller the spacing.

Vector overlay. For conductors, the electric field lines must be perpendicular to the surface.

1.The spherical conductor of the cavity, in the absence of an external electric field, will have an even charge distributed on the outer surface of the spherical shell. The electric field in the cavity is zero. If it is not a conductor, the charge distribution is constant, determined by the initial state, and can be uneven.

2.The drawing method of equipotential surface is only a representation, for example, drawing equipotential surface at intervals of 1v, obtained by v=e*d, the larger e, the smaller d.

3.Electric field lines are issued (positive charge) or aggregated (negative) from the star at the charge, and the greater the amount of charge, the more electric field lines. The equipotential surface is perpendicular to the electric field lines. When there are two charges, the equipotential surface is generally bent towards the side with the greater amount of charge (absolute value). (Didn't verify it carefully.) ）

4.For a conductor in electrostatic equilibrium, the electric field strength on the surface, if present, must be perpendicular to the surface. If there is a potential difference between the inside and outside of the surface, the value is not zero. The conductor is equipotential and the outside world is not, and the electric field on the surface of the four parallel planes will not be zero.

1.If it is a spherical conductor with a cavity inside, if it is in electrostatic equilibrium, how is the charge distributed? What is the electric field in the cavity? What if the sphere is not a conductor?

2.Why is the greater the field strength, the smaller the distance between the two equipotential surfaces?

3.How to draw a diagram of electric field lines, equipotential surfaces? (especially the combined electric field of two point charges with different magnitude of charge).

4.If a conductor in electrostatic equilibrium is a regular cuboid, the long line of the external electric field is perpendicular to the two surfaces, is there four surfaces (i.e., faces parallel to the external electric field line) that have zero electric field strength?

Or is the field strength line perpendicular to its surface?

The option is not greater than less than or equal to l 2.

Let the original charge be q, and analyze the force of one of the balls, and balance the tension of the rope, the re-shaped force and the Coulomb force. Let the angle between the rope and the vertical direction of Zina be a, then mgtana=kq 2 l 2

If the charge of the ball is reduced by half, the distance between the two balls will be shortened, and the angle between the rope and the vertical direction is BL 2

If a positive charge is introduced, then all the forces are repulsive and cannot be balanced, if a negative charge is introduced, consider them on both sides and in the middle of the two charged balls, and the analysis shows that it is possible to equilibrium only if they are placed in the middle of the two charged balls, and then it is a column calculation, according to the formula f=kq1q2 r 2 and the sum of the forces of the other two balls on any ball is 0

You can solve the equation by 3 equations (the distance between the position and the +q ball is r, and the charge of the ball is q3).

Negatively charged, placed on the connecting line of the previous two, at a distance of 9q, at a distance of q.

The power is 9q 16

Idea: First of all, it is certain that it should be negatively charged and should be placed on the connecting line or extension line of two small balls with dots, specify a positive direction (such as a positive direction to the right), assume that the distance from +q is "x", and then arbitrarily select a charged ball to solve it with equations according to the force balance relationship.

The reason why the rope has no tension when the charge is released is because the component of gravity along the direction of the rope and the electrostatic force are just equal.

mgcos60=kq1q2/r^2

Bring in the data to find the power of A.

The problem does not need to be supplemented, the number of parallel pairs does not change, and if it changes, it means that it is not parallel.

This topic is to test the internal resistance of a voltmeter, under normal circumstances, we do not consider the internal resistance of the voltmeter, so some students are not clear. The composition of the voltmeter is actually made up of a galvanometer and a large resistor in series.

The voltmeter and the resistor are connected in parallel, and the parallel circuit resistance r=r1r2 (r1+r2).

V1 V2 measures R1 R2 respectively, and the voltage of the series circuit is proportional to the resistance of the 35V series, then there is R1R1 (R1+R1)=R2R2 (R2+R2).

V1 V2 is reversed, then V1=30V, V2=40V R1R2 (R1+R2):R2R1 (R2+R1)=30:40

Substituting r1=1800 r2=1200 gives the simultaneous solution r1=5760 r2=9600

Since R1R2 is in series without considering the internal resistance of the power supply, then there is: R1 and R1 = R2 and R2

After the modulation R1 and R2) (R2 and R1)=40 30R1=1800 ,R2=1200, R1 is V1 internal resistance, R2 is V2 internal resistance) can be calculated!

The question stem is incomplete, and this expression can't be solved.

Are r1, r2 the same circuit or a different circuit?

If it is located in circuit 1 and circuit 2 respectively, and these two circuits have other resistance and power supply factors (considering similar practical use), it should not be solved.

Think of the voltmeter and its corresponding resistors as a new resistor, i.e., the two resistors are connected in series, i is the same: r1 r2=v1 v2

r1×r1/(r1+r1) :r2×r2/(r2+r2)=35 : 35

r2×r1/(r2+r1) :r1×r2/(r1+r2)=30 : 40

Substituting r1 and r2 yields r1=5760 euros and r2=9600 euros.

Are R1 and R2 in series or parallel?

f=bilcos is obtained by f=bilcos

The number of protons at 0 is the determination of the type of matter, and generally does not change, and the nuclear cavity or the free electrons that change when the electric conductor and the non-electric conductor come into contact with each other are not protons, so the number of protons does not change, and c is chosen

1, f=bilcos is obtained by f=bilcos

After the uncharged conductor A and the negatively charged conductor B come into contact, the negatively charged object attracts the mass beat of A, and the proton decreases until equilibrium. Pick B

1) At the beginning, there is: f=ma=kqaqb l2 for b, and after a period of time, there is: f=2ma=kqaqb r2 and the old skin and the hungry world are solved in two formulas: r = root number 2 2l =

2) Conservation of momentum: MVA=2MV

Va=2V, the difference in direction is opposite to b.