On the topic of field strength, about the concept of field strength, the following statement is corr

Don't you know the direction of gravity (the direction of gravity of the ball is set to the left) add: Since ball A is on the left, then their electric field force and gravitational force are exactly the opposite of mine.

1.First of all, the two balls have a dissimilar charge, so there is a gravitational force between them, then the direction of the electric field is opposite to the direction of their gravitational force.

For the analysis of the force of the two small balls, the small ball with a positive point is subjected to the upward pull of the rope, the downward gravity, and the gravitational force on the left, because they are stationary, the electric field force is of course to the right, (in the same way, the direction of the force of the negatively charged ball should be known to you) according to Coulomb's theorem, the Coulomb force is the electric field force, so you know the electric field strength.

Just read the wrong question) at the midpoint of the electric field field is strong and directional.

i.e. their sum field strength The direction of the electric field of the ball a is directed by a towards b

The direction of the electric field of the ball b is also to the right, so the combined field strength is to the right.

The size is directly with the formula, the formula of the field strength.

a. Just a calculation formula, just like in the formula r=u i, the resistance is not directly related to the voltage and current.

b. The field strength has no relationship with the positive and negative charges put in, the positive charge is forced in the electric field in the direction of the electric field, and the negative charge is the opposite.

c. Correct. d. Is the field strength objectively existing and has nothing to do with the test charge, or is it an example in a, can it be said that the resistance is 0 without energizing the resistor?

The field strength at each point in the electric field is fixed and does not change depending on the charge (including the mass, amount of charge, and positive or negative charges)!

2l, because d does not change. So y does not change, because the electric field does not change, so a does not change, from y=1 2at 2 t does not change, because the horizontal displacement x=vt, so when v becomes 2v, x also becomes twice as much.

Let the ball on the left be positively charged.

The right side is negatively charged.

Balanced forces. The ball on the left is subjected to the gravitational pull of a negative charge.

kq^2/r^2

Direction to the right. It is balanced by the force of the electric field.

Hence the direction of the electric field is from right to left.

qe=kq^2/r^2

e=kq/r^2

Substitute the numerical value to solve it.

;It's okay not to explain.

At the moment of release, the system obtains acceleration (6mg+qe) 6m=g+qe 6m, for a, a is subjected to gravity mg and tensile force t ab, according to Newton's third law, m(g+qe 6m)=mg+t ab, t ab=qe 6

2l, because d does not change. Therefore, y does not change, because the electric field does not change, so a does not change, from y=1 to 2at 2 to get t unchanged, because the horizontal displacement x=vt, so when v becomes 2v, the world object search x also becomes twice as much.

Solution: When the initial velocity of the charged particle is v, it is obtained from the law of flat-like motion:

d=12at2=12a(lv)2

The initial velocity of the charged osmotic particle becomes 2V, the distance of the deflection of the particle from the electric field is y, and the deflection angle of the velocity of the cong source is

Then we get: y=12at 2=12a(l2v)2 from the above two equations: y=14d

According to the inference, the reverse extension of the velocity of the particle when it leaves the electric field intersects at the midpoint of the upper plate according to the geometric knowledge.

tan = yl2 = d4l2 = d2l, and blind spring tan = dlab solution gets: lab

2l so x=