Sophomore physics electrical problems, sophomore physics problems electricity

1) The question is that the power supply is connected to the two boards, and the voltage of the two boards remains unchanged.

It turns out that the particles hit the n-plate**. Let the plate length be l and the muzzle velocity of the particle when it enters the electric field is v , then.

a qe m q u (m d ) u is the supply voltage, q is the particle charge, m is the particle mass).

l / 2＝v* t0

d ＝a*t0^2 / 2

The above three forms of synapism d [ q u (m d )]l 2 v) 2 2

d root number [q*u*l2 (8 m*v2)

If you want the particles to fly just out of the electric field (the power supply is still connected to the two plates), let the distance between the two plates be d1

then a1 qe1 m q u (m d1 ).

l ＝v* t1

d1 ＝a*t1^2 / 2

The above three forms of Synapid d1 [ q u (m d1 )]l v) 2 2

D1 root number [ q*u*l 2 (2 m*v2 ) 2 d

So the distance that the n-plate should move down is.

d1 d 2d d d (indeed down distance d).

2) When disconnected, the electric field strength between the two plates does not change (the plate charge also does not change, it is not proved here, if necessary, it is proved separately).

Let the distance between the two plates be d2, and the particles will fly out of the electric field.

a2＝qe / m＝q u2 / (m d2 )＝q u / (m d )

l ＝v* t2

d2 ＝a2*t2^2 / 2

The above three forms are synapoid d2 [ q u (m d )]l v) 2 2

Compare the d root number [ q*u*l 2 (8 m*v 2 ) with the above equation to get .

d2＝4 d

So the distance that the n board should move down is d2 d 3d

The horizontal velocity does not change, and the time is twice as long as the horizontal distance is to make it twice as long, and here the electric field force and the vertical distance are changed due to the increase of d.

Suppose the spacing becomes d'

The original electric field force, vertical acceleration a=uq dm

Now it becomes a'=uq d'm

a’=a(d/d')

d=at²/2

d'=a(d/d')·(2t)²/2

d‘=4(at²/2)(d/d')=4(d/d')dd'=2d, so the plates should be moved up d'-d=d

2.When the power is turned on, the capacitor does not change the voltage, but when the power supply is disconnected, the capacitor does not change the amount of charge.

And the voltage of the two pole plates of the capacitor is u=q c

c 1 d so u d

Acceleration a=uq dm, i.e. a and u are independent.

So the vertical acceleration is constant.

Set the spacing d after the change''

t''=2t (the first step has already stated the reason).

So d''=4d

It should be moved down 3d

The first question is d, and here I will talk about the second question: when the switch is disconnected, the field strength between the two plates does not change, so the electric field force does not change before and after the board moves. Let the vertical acceleration be a.

Suppose the first time on the n-board ** used time t, then the second time must be 2t (horizontal direction of uniform motion), from the title: d=a*t squared 2 (vertical direction of the initial velocity of zero uniform acceleration). Then after the plate is moved down, there is l=a*(2t) squared 2, (Note:

This l is the total distance between the two plates) to get l=4d, l-d=3d.

Summary: For the problem of uniform acceleration with uniform velocity in one direction and zero initial velocity in one direction (zero initial velocity is very important, don't use it indiscriminately), if the points are taken at equal intervals in the direction of uniform velocity, the distance ratio of two adjacent points in the direction of uniform acceleration is 1:3:

This question is 3D. Reason: The distance in the direction of uniform velocity is equal to the same time, and the distance ratio in the direction of uniform acceleration is changed in square from the beginning, that is:

1 square: 2 square: 3 square:

4 square. The above 4d is 2 squares) then the distance ratio of adjacent 2 points is the latter minus the former, 1 square: (2 squares - 1 square):

3 sq - 2 sq) :( 4 sq - 3 sq) ...Using the squared difference formula is equivalent to:

4+3）..Seeing the difference between the two adjacent numbers is 2, and finally we get 1:3:

In this way, it is much simpler to encounter problems such as flat throwing, for example, if you draw 5 equally spaced grids in the horizontal direction, and ask you the distance ratio between the first grid and the fifth grid in the vertical direction, obviously you don't need to calculate, it is 1:9

Question 1b

To the left of Q1, because Q1 Q2 and the length R1 from Q1 is smaller than the length R2 from Q2, Q1 is larger and closer, so the field strength of Q1 is greater than that of Q2, so the field strength cannot be 0

The middle part of the electric field lines is unanimously to the right, and the field strength cannot be 0

On the far right, in this straight line, there is a point that satisfies the field strength of 0, because the length of the right part of the distance from q1 r1 is greater than the length r2 from q2, although q1 is large, but the distance is farther, and there is a point that satisfies a certain relationship between distance and charge so that the field strength is 0

It shows that there is only one point that satisfies the combined field strength of 0.

Find the point of 2e2 below.

On the far left, because q1 q2, r1e2, no point can be equal to 2e2

In the middle, since the middle part r1 gradually becomes larger from left to right, r2 gradually becomes smaller, so e1 gradually becomes smaller and e2 becomes larger, so there must be a little satisfaction in the middle e1=e2

At this point, the field strength is exactly equal to 2e2.

Since the field strength of Q2 decreases immediately from infinity to the right, the field strength of Q1 decreases at a slower rate than that of Q2 because the distance is already relatively far.

So there is exactly a point where e2 can be reduced to exactly equal to e1, but in this case, the sum field strength is 0 because of the opposite direction

So there is only one point at 2e2.

In the second question, the direction of the electric field at A is to the right according to the electric field line, and B is also to the right.

B is wrong, there is no such statement at all, the size of e is only determined by the density of the electric field lines.

c False, ditto.

D: Yes, if you can tell me the distribution of the electric field lines at points A and B, which is the degree of density, I can judge the size of EA and EB.

1, the electric field strength e=kq1q2 r, this line is divided into three parts: left, middle and right, the left is because q1>q2, r1e2. The other two parts, q1>q2, r1>r2, so each has a little e1=e2, and the electric field in the middle must be to the right, so this point e=2e2.

2. The electric field strength is only related to the density of the electric field line, only one is known here, and I don't know how the other is distributed, so I don't know the density, and I don't know the intensity, and the electric field line points to the direction of the electric field, so choose AD.

(1) e=kq r 2 because q1=2q2 if e1=e2 then e1=kq1 r 2=kq2 r 2=e2 so r = root number 2 times r

Because the field generated by the positive charge is outward, and the field generated by the negative charge is inward, both are radial, so there are two points that match such a point, that is, between q1 and q2, and to the right of q2. Judging by the direction of the electric field lines, the one between 2e2 and the one on the right is 0 on the left.

2) The direction of the electric field strength is along a straight line (the electric field strength of the curve is the tangent of the curve), so a pair. There is no certain relationship between the electric field line and the field strength, for example, the field strength in the uniform electric field is equal everywhere, so B is wrong. In the same way, c is also not true, for example, the field strength is not equal when the field is generated by a point charge.

So d is correct.

There is an r in the formula e=kq r 2, which refers to the distance between the tentative charge and the field (source) charge, which is specifically referred to as the electric field of the charge. Since there is no so-called center of electric field in an electric field other than the electric field of the point charge, Equation 2 is obviously a bit inappropriate in this case.

r has already been explained in the previous paragraph. In fact, the second formula tells us that the strength of the field at a certain point in the electric field of the point charge is only related to the charge itself and its position, and not to the field source charge. The position of the test charge mentioned above is reflected in the formula as the variable r, which represents the distance from the field charge.

It is according to the first formula that the magnitude of the electric field is defined, so it certainly applies to any electric field. The second formula is a formula that is specially suitable for finding the electric field of point charge after the first formula is obtained by using the method of inductive hypothesis and obtained a large number of experimental verifications, so it is only applicable to point charge.

r refers to the magnitude of the electric field strength at the distance point charge r.

Small bulb, labeled "Working i=p ul=

Parallel circuits, one with a light bulb and one with an electric motor.

The voltage at both ends of the motor ud=

Ohm's law by closing circuits.

e=ul+u' u'=1v i=u'r=current through the motor id=i-il=

iud=id^2r0+p

p = output power of the motor.

Supply Current: (

Total power: motor power:

This is easy to understand, the direction of the arrow represents the direction of the electric field strength, and the charge in the problem has a negative charge, and the direction of the electric field force is opposite to the direction of the electric field strength, so the negative charge at rest.

Put in it to get from A to B, the arrow must be from B to A, and the charge can be subjected to the force that makes it.

Movement from A to B . And its acceleration is getting bigger and bigger, which means that it is experiencing more and more force, that is.

The electric field strength is getting bigger and bigger, so the electric field lines are getting denser and denser, so C is chosen

The main thing is diagram C: Why is the arrow pointing towards A and not towards B like the D diagram?

Because it's a negative charge ...

The negative charge naturally moves towards a place with a high electric potential, and the electric potential of b is judged by the density of the electric field lines.

Judging from the velocity time graph, the acceleration of the negative charge is constantly increasing, and Newton's second law indicates that the force is getting bigger and bigger, and it must be that the closer the charge from the field source, the greater the force

From the V-T image, it can be seen that the speed of the charge is getting bigger and bigger, and the acceleration is getting bigger and bigger, so it can be judged that the electric field force it is subjected to is getting bigger and bigger, and it can be seen that the electric field strength from A to B is getting bigger and bigger, that is, the electric field line is getting denser and denser, and at the same time, because it is a negative charge, the direction of the electric field is roughly opposite to the direction of AB, so C is chosen.

Because option ABC is wrong, there is no answer to the question you asked, and if you can know that option D is correct, it's OK.

As can be seen from the V-T image, A is an accelerated motion, indicating that the force (electrostatic force) experienced by A is increasing, so A is moving in the direction of high density of power lines. Accordingly, items A and B are excluded;

and A moves from rest and is only affected by electrostatic force, so A must move in the direction of positive charge, that is, the direction of the power line is from B to A.

In summary, it can be seen that the answer is item C.

From the velocity diagram line, it can be seen that the electric field force of the "negative" charge increases from A to B, and accelerates, so the field strength at point A is less than the field strength at point B, and the potential at point A is lower than the potential at point B.

Hello! Electric potential u=e electric potential energy q

In the electric field, the ratio of the potential energy of a charge at a certain point to the amount of charge it carries is called the potential of this point, and the potential energy of a point in the electric field is the point where the charge moves to the 0 potential energy point, the work done by the electric field w We know the distribution of the electric field in the middle of two equal heterogeneous charges, the direction of the electric field at the midpoint is the horizontal direction of the two charges, therefore, if the charge is moved on this vertical line to the 0 potential energy point at infinity, the work done by the electric field force is 0, so w = 0, that is, the electric potential energy is 0, so the electric potential = 0

I don't know if I understand o( o ha!

The electric potential formula is kq, r is the distance from one point to the point charge, the electric potential is a scalar quantity, the distance from the midpoint to the two charges is r, the charge is equal to the other, and the sum of one positive and one negative is 0

If you don't understand, ask me again.

Assuming that both dissimilar charges are at infinity at the point of zero potential, then the potential formed by the positive charge is positive and inversely proportional to the distance; The potential formed by a negative charge is negative and is also inversely proportional to the distance. Note that at this time, at the point of absolute zero electric potential at infinity, the electric potential formed by the two at the midpoint of l at infinity is equal but positive and negative is opposite, and the neutralization is zero!

The electric field formed by the vacuum point charge e kq r2 , according to the principle of superposition of electric fields, the electric field at the middle point of the two charges must be zero.

The electric potential is different from the electric field strength, the electric potential is numerically equal to the work done by the electrostatic field force when the unit positive experimental charge is moved from a certain point to infinity, and the electric potential of a point in the electric field excited by the point charge system is equal to the algebraic sum of the electric potential established at that point when the charge at each point exists alone, the formula v=1 (4 e)*q d, and the problem is two different charges, then the value of q is reversed, so the sum of the two is equal to 0