Physics, Question 20 on Elementary Electricity in High School

First of all, the first two point charges are each subjected to only one force, so to achieve equilibrium with the addition of a third charge, the third charge must be on the same level as these two charges.

Then, we consider where the third charge is placed.

Suppose placed between the first two charges, neither positive nor negative charges can balance the three charges. You can try to draw a diagram and analyze the force.

Therefore, it can only be placed on a straight line outside of the two charges. Because the positive charge has a larger charge and therefore a greater force, the third charge should be close to the negative charge, that is, next to the negative charge. Q1 - Q2 - Q3.

It is clear that a positive charge should be put on it. Then calculate the distance of q2 q3 when the equilibrium is reached;

Two unknowns, the equation of the equilibrium of the two charges can be solved. q3=45×10^-3 l=50cm

The triangle rule can be calculated, but it feels a little difficult to calculate, hehe!

First of all, I would like to declare that I am not very good at philology.

1.Analyze the force: the third charge must be on the line of the two charges known; Moreover, it must no longer be in the middle.

2.Overall analysis: The absolute value of the negative charge is large, so the third charge should be at the distance from the positive charge.

f=kq1q2/r2)

3.Calculation: Calculate the column equation by yourself. (Q3 is definitely negatively charged, I don't need to say this).

BC is chosen, which is done by using the electric field force equal to the centripetal force, and because it is a gap, it can be considered that the radius of motion of the particle r has always been constant.

Use the formula eq=mv 2 r

a.When the amount of charge is the same, the right side of the equation is also the same, and since the velocity v is not known, it is impossible to judge whether the mass m is the same.

b.When the amount of charge is the same, the right side of the equation is also the same, the kinetic energy is 1 2mv 2 so it is the same, b is correct.

c.If the ratio of the amount of charge to the mass of the incident particle is equal, i.e., eq m=v 2 r, it can be judged that v is the same, and c is correct.

d.If the ratio of the charge to the mass of the incident particle is equal, i.e., eq m=v 2 r, it can only be judged that v is the same, and the kinetic energy is 1 2mv 2, because the mass m does not know, it is impossible to judge whether the kinetic energy is the same.

Analysis: Answer: B c

Charged particles enter the electric field, and the electric field force acts as a centripetal force, which is obtained from Newton's second law:

qe=mv2 r yields:

r=mv^2/(qe)

According to the flow of positively charged particles from the electric field region m at one end of the electric field region, along the semicircular orbit shown in the figure, through the electric field and from the other end n can be seen: r is a certain value.

Analyzing the a, b, c, and d options, only bc satisfies r is a certain value. Therefore, BC was chosen.

I hope you are satisfied, and you can chat secretly if you have any questions.

AD analysis: When the velocity is the maximum when B is reached, that is, the acceleration is 0, that is, the electric field force is equal to the gravitational force, and the field strength at point B can be calculated.

When the velocity is 0 when it reaches point C, it is conserved by energy, and all the potential energy is converted into gravitational potential energy, so the electric potential at point C can be found!

As for why we know why we can't find the field strength at point B, and why we can't find the field strength at point C, the main reason is that the distribution of the electric field generated by the charged disk is not very clear, I think it should be oblate spherical, so it is difficult to determine how the relationship between the electric potential and the field strength is expressed by the distance h!

B, the energy is conserved in the motion of the ball and the kinetic energy at point O and point C are both 0 so the potential energy at point C is equal to point O.

ac.Since b has the largest velocity, the electric field force is equal to gravity, and both the potential of b and the field strength of b can be known.

a, under the joint action of electric field force and gravity, the ball does a variable acceleration motion with smaller and smaller acceleration upwards, and the velocity is the largest at point b, which means that the resultant force at point b is zero, that is, eq=mg, and the m and q of the ball are known in the problem, so the field strength of point b can be obtained.

The capacitor measures the road-end voltage at both ends of the power supply UAB=U=E-E (R12+R3) When the sliding contact of R1 slides to the right, the resistance of R1 becomes larger, the parallel resistance of R12 becomes larger, the total resistance of the external circuit becomes larger, the current of the external circuit decreases, and the number of current representations decreases. The road-end voltage increases. As the UAB increases, the electric field strength between AB increases, and the electric field force on the charged oil droplets increases.

To move upwards. A false. b correct.

Only the distance between the two plates A and B is increased, the current is unchanged, the UAB is unchanged, the electric field strength between AB is weakened, the electric field force of the oil droplets is reduced, and the oil droplets move downward. c correct.

Only the relative area between the two plates A and B is reduced, the UAB remains unchanged, the electric field strength between AB remains unchanged, and the oil droplets remain balanced and do not move. D false.

Capacitor voltage = external circuit voltage, ammeter reading = circuit current, oil droplets between capacitor plates are subjected to downward gravity and upward electric field force, and the two forces are balanced.

The R1 contact to the right causes the access resistance of R1 to increase, the total resistance value to increase, the current to decrease, the internal circuit voltage to decrease, the external circuit voltage to increase, the voltage at both ends of the capacitor to increase, the field strength to increase, the oil droplet to receive the electric field force to increase, and move upward.

Increasing the spacing between the two plates of the capacitor will not change the total resistance of the circuit, so the current and voltage are unchanged, but the increase of the spacing between the plates will lead to a decrease in the field strength, and the oil droplets will be reduced by the electric field force and move downward.

Changing the positive area of the capacitor only changes the capacitance of the capacitor, and has no effect on the electric field, and the force of the oil droplets remains unchanged and does not move.

So choose BC

B should be selected - the external resistance becomes larger, the total current becomes smaller A indicates becomes smaller, the terminal voltage becomes larger, the plate spacing remains unchanged, the field strength becomes larger - the gravity is less than the field force - upward movement.

c - the change of capacitance does not affect the circuit current, a indicates the same as the indicator, but the field strength becomes smaller - the gravity is greater than the field force - downward motion.

B&cb does not need to be explained, the overall external resistance becomes larger, i decreases, and u increases.

c Think about it, if you say that the plates are infinitely expanded, the electric field between the two plates will disappear and the oil will fall.

Electric field lines and equipotential surfaces.

The projection of the line segment ab in the direction of the electric field line d=ab*cos(90°-53°)=

e=uab/d=|(-10)-10|/

The field strength of the uniform electric field is e=U d=20Vlsin53°=1000Vm

In this case, the change in u2 is equal to the sum of the change in the internal voltage and the change in u1.

Because U1 i is equal to R1, it does not change, so U1 I does not change, therefore, a is correct U2 I is the resistance of the sliding rheostat, which becomes larger, but the value of U2 i is actually the sum of the internal resistance and R1, because the electromotive force of the power supply does not change, therefore, the voltage on the sliding rheostat becomes larger, which is equal to, the sum of the internal voltage and the amount of power reduction on R1, so B is wrong.

So, C is correct.

U3 i is equal to the end resistance, so it becomes larger, while u3 i is actually equal to the internal resistance, so it does not change, d is correct.

U is the same as U1 is positive or negative, the sum of the two is the opposite of U2, and the equal size of U3 i=r u2 i=r+r1 u1 i=r1

I don't know if I want to ask later.

Analysis: From the circuit diagram, it can be seen that R1 and R2 are connected in series, V1 measures the voltage at both ends of R1, V2 measures the voltage at both ends of R2, V3 measures the voltage at the end of the road, and the change of voltage and current can be obtained from Ohm's law. The ratio of the change in voltage to the change in current can be obtained by Ohm's law of a closed circuit

Answer: Solution: Since r1 is constant, then Ohm's law can be obtained, and u1 i=r1 remains unchanged; From the mathematical law, u2 i also remains the same, so a is correct; Because the slide moves downward, the access resistance of R2 increases, so U2 I becomes larger; And u2 i= i(r1+r) i=r1+r, so u2 i remains the same, so b is wrong and c is correct;

Because u3 i=r1+r2, the ratio increases; And u3 i= ir i=r, so the ratio remains the same, so d is correct;

Therefore, ACD is chosen

Comments: The difficulty of this question lies in the analysis and judgment of the ratio of the change of U2 and U3 voltage to the change of current, and the answer to B can no longer be simply derived from A, when the resistance changes, the ratio of the change of voltage and current no longer remains unchanged, and the expression of the voltage change value should be obtained through the analysis of Ohm's law of the closed circuit, and then analyzed

If you have any questions, you can ask them.

δu1 δi denotes the resistance of r1 r1 is a fixed value resistance, so this ratio is constant. A correct.

U2 δi is the sum of R1 and the resistance in the power supply, which is a fixed value, so this ratio does not change, B error C is correct.

U3 δi represents the resistance within the power supply and is also a fixed value. So this ratio also does not change, d correct answer acd

Because after releasing the ball, the ball is subjected to an electric field force to the right, a downward gravitational force and a pull force of the rope, where the electric field force and gravity do the work.

The electric field force does negative work, so the electric field can increase, and gravity does positive work, so the gravitational potential energy decreases.

Assuming that the ball can fall to the vertical direction of the string, then the work done by gravity and the electric field at this time is calculated blindly.

w(weight)=mgl= j w(electric field)=eql= j Since the work done by gravity is greater than the negative work done by the electric field force, the ball still has a certain amount of kinetic energy at this time, so the ball can pass through the lowest point.

The kinetic energy is w=w(weight)- w(electric field)=3x10 (-3) j kinetic energy w= 0 5mv 0 5, then v= (2wm)= 5 ms

Electric field acceleration qu0 = mv0 2 2

In the deflection electric field: l=v0t

When the deflected electric field is ejected, the partial velocity in the direction of the vertical initial velocity vy=at=qel mv0

Then the deflection angle satisfies: tan = vy v0 = qel mv0 2 = el 2u0

1) If they are accelerated by the same electric field and then enter the same deflection electric field, then tan 1:tan 2 = 1:1

2) If they are entering the deflected electric field at the same velocity.

According to Tan = vy v0 = Qel mv0 2 we know that Tan 1: Tan 2 is proportional to the specific charge q m of the particle, i.e.

tanφ1:tanφ2=1:2

3) If they are entering the deflection electric field with the same momentum.

According to Tan = vy v0 = QEL MV0 2 it can be seen that Tan 1: Tan 2 is proportional to the amount of charge of the particle, i.e.

tanφ1:tanφ2=4:1

The first one has to do with the amount of electricity of atoms and protons, and the amount of electricity is different, and the speed is different. The second is related to power, and the third is related to quality!