A simple question about electric charges

Solution: The first question is simple, but it seems that none of the upstairs are completely correct.

Ask whether the direction of the electric field at point A should be taken into account?

Because point A is to the right of Q, there is a negative charge, and because it is subjected to the electric field force to the right, it can be deduced that Q is a positive charge.

Because the direction of the electric field is the same as the direction of the electric field force experienced by the positive charge in the electric field, the magnitude of the electric field is and the direction is horizontal to the left.

In (1), it has been proved that q has a positive charge.

Already f=,q= c,r= m,k=9*10 9 n*m2 c 2q=f*r 2 kq

2*10^-7 c

From the formula e=kq r 2, it can be seen that when k and q are constant, the larger r is, the smaller e is.

When r takes the maximum r(max)=, e has a minimum value e(min).

e(min)=kq/r(max)^2

9*10^9 n*m^2/c^2 )*2*10^-7 c )/(n/c

The above results are completely in accordance with oral arithmetic and mental arithmetic, and it is inevitable that there will be errors and omissions, I hope you will correct them, thank you.

e=f/q=

Because the negative charge is subjected to a horizontal to the right force in this electric field, the two point charges are attracted to each other, so q is positively charged.

q= ce=kq/r2=

Upstairs, don't bring a negative sign when calculating Q, the result is to be wrong!!

ea=f q=2*10-5 -10-9=2*104(n c)q, positively charged, and the two attract each other.

k=9*10^9）

e=kq r2=slightly (bring in Q).

a. A special substance that exists in the space around the charge, which is the electric field, so a is correct;

b. The electric field is a special substance that exists in the space around the charge, and the electric field is different from the usual physical object, but it exists objectively, so B is correct;

c. The nature of the force of the electric field is that the electric field has a force on the charge placed in it, and this force is called the electric field force, so c is correct;

d. The electric field is a special substance that exists in the space around the charge, and the electric field is different from the usual physical object, it exists objectively, so D is wrong;

Therefore, ABC is chosen

Two different ways to power up this time.

Duqian jujube can be understood directly from the name.

For example, when glass rods, silk, etc. rub against each other, electrons are transferred, so that two objects are charged with the same amount of dissimilarity. The direction of electron transfer can be judged according to the change in the charge of the object after friction.

Object A is charged and touches the uncharged object B. If A is positively charged, then the electrons on B are transferred to A, so that Xiao B is positively charged; If A is negatively charged, the electrons on A are transferred to B, making B negatively charged.

This is a trajectory of a charged particle with an initial velocity in an electric field.

If it is seen as a circle, then the center of the circle is on the left side of the trajectory, which is opposite to the direction of the electric field, so it has a negative charge, which is the meaning of "inward", if it is a positive charge, it should turn to the upper right corner, and the center of the circle should be on the right side of the trajectory.

Since it is a negative charge, and it is outward (far away) along the electric field line, the electric field does negative work, so the above conclusion that the velocity is constantly decreasing is drawn when only the electric field force is applied, and if there are other forces, the situation is different.

There is an electric field around the charge, independent of the number of charges. So, there is an electric charge and there is an electric field.

If another charge is placed around a charge, the field strength around the original charge will definitely change, and it is calculated according to the principle of superposition of field strength.

To study the electric field strength of a certain point around the charge, when using the test charge for experiments, it is necessary to try to avoid the influence of the test charge on the electric field of the original charge due to the placement of the test charge, which requires that the test charge should be small and small in volume compared to the original charge.

Of course not, there is only one charge, if it is a positive charge, the direction of the electric field is from the position of the point charge to infinity, and the negative charge is from infinity to the point charge, and the absence of field strength means that several field strengths at this position must be offset to zero.

The electric field force of the ball remains constant, and the electric = eq. Gravity remains the same, g=mg

When you reach B, do a circular motion. The centripetal force required to be f=mv2 r is provided by the component of the supporting force of the ring. Therefore, what you need to look for is the speed size.

From A to B, the work done by gravity is MGR and the work done by electric field force is EQR. All converted into kinetic energy of the ball 1 2mv2

v = root number [2(g+eq m)r]. The magnitude of the centripetal force is 2mg+2eq.

At this point, after the supporting force of the ring is decomposed, G is to be cancelled out, F electricity is to be cancelled out, and then the centripetal force is also provided. So the size is.

Root number [(mg) 2+(2mg+2eq+eq) 2] = root number [5(mg) 2+12mgeq+9(eq) 2].

The ball slides to point C and can be analyzed in the same way.