Point charge field strength formula, r approximates with 0 when E approaches with infinity why

In this case, the formula will no longer apply, because the infinite approach is already at the micro level, and it cannot be explained by this macro formula.

In a uniform electric field: e=u d.

The field strength of a uniform electric field e = uab d {uab:ab voltage between two points (v), d:ab distance between two points in the direction of field strength (m).

If you know the magnitude of the force on a charge, the strength of the electric field.

It can be expressed as: e=f q.

The electric field formed by the point charge: e=kq r 2, k is a constant, q is the amount of charge of the charge, r is the distance to this charge, it can be seen that with the increase of r, the field strength formed by the point charge gradually decreases (the field strength formed by the point charge is inversely proportional to r 2.

Precautions

1) When two identical charged metal balls are in contact, the power distribution law is as follows: the original dissimilar charge is neutralized first and then equally divided, and the total amount of the original band of the same charge is equally divided.

2) Electric field lines.

Starting from a positive charge and terminating with a negative charge, the electric field lines do not intersect, tangent.

The direction is the direction of field strength, the field is strong at the dense part of the electric field line, and the potential is getting lower and lower along the electric field line, and the electric field line is perpendicular to the contour line.

3) The distribution of electric field lines for common electric fields requires memorization.

4) Electric field strength (vector) and electric potential (scalar.

are all determined by the electric field itself, and the electric field force.

It is also related to the amount of electricity in the charged body and the positive and negative charge.

5) The conductor in electrostatic equilibrium is an equipotential body, the surface is an equipotential surface, the electric field line near the outer surface of the conductor is perpendicular to the surface of the conductor, the combined field strength inside the conductor is zero, there is no net charge inside the conductor, and the net charge is only distributed on the outer surface of the conductor.

6) Capacitance unit conversion: 1F=10 6F=10 12Pf.

7) Electronic volts.

ev) is a unit of energy, 1ev=.

and its applications, equipotential surfaces, tip discharges, etc.

9) The electric field strength e=u d=4 kq s, and the work done w=u*q.

No, when r approaches 0, the shape and size of the charge can no longer be ignored, and then the charge is no longer a point charge, because the shape and size of the charge can no longer be ignored.

The field strength of a uniform electric field is e=uab d.

Electric field force: f=q*e.

Potential to potential difference: UAB = a- b, UAB = wab q = -δeab q.

The work done by the electric field force: wab=q*uab=eq*d.

Electric field: 1Two kinds of charges, the law of conservation of charge, and the element charge: (e=; The amount of charged body charge is equal to an integer multiple of the element charge.

3.Electric field strength: e=f q (definition, calculation, field strength is its own property and has nothing to do with electric field force and electricity).

E: electric field strength (n c), which is the vector (principle of superposition of electric fields), q: the amount of electricity to be tested (c)}.

4.The electric field formed by the vacuum point (source) charge e=kq r2.

This is not right ... Our teacher has mentioned that the formula will no longer be applicable in this case, because it has already reached the micro level when it is infinitely close, and it cannot be explained by this macro formula.

Suppose a sphere is cut in the middle to become two hemispheres on the left and right, and the right hemisphere is selected.

A uniform electric field with a field strength of e.

Neutralize the empty macro, assuming that the deficit e is to the right.

then the strength of the electric field through this hemispherical surface.

Flux = electric field strength flux through the plane on the left side of the hemisphere = s circle * e = *r * r * e

According to the sum of kinetic energy and potential energy is the conserved limb, because on the equipotential surface 2 the electric year is 0, the electric potential energy is also 0, and its kinetic energy is half of the sum of the two kinetic energies of the equipotential surface 1 and the equipotential surface 3, that is, it is 10 joules, so on each equipotential surface, the sum of kinetic energy and potential energy is 10 joules, so when the electric potential energy is 4, the kinetic energy is 6 joules.

Well, the first sentence is for the blind because of the initial disadvantage of the point charge with zero velocity.

Only in the electric field force.

When the reed is empty under the action, the electric field force does positive work on it and the potential energy decreases.

The second sentence is false A negative charge with zero muzzle velocity moves from a low electric potential to a high electric potential only under the action of an electric field force.

That's it.

a. The point charge field strength formula is based on e=f

q derivation, where f is the Coulomb force of the field source charge to the tentative charge When r tends to 0, the Coulomb force formula f=kqq does not apply

r medium distance requirements, so a is wrong

B. Each quantity in the formula ep = q should be substituted into the symbol, and it can be seen that the negative charge has a large potential energy in the place where the electric potential is low, so B is correct

C. Capacitance C is determined by the capacitor itself, which is not proportional to the amount of charge Q and inversely proportional to the voltage, so C is wrong

d. D in the formula UAB=ed is the distance along the direction of the electric field, so the greater the distance between any two points A and B in the uniform electric field, the greater the potential difference between the two points, so D is wrong

Therefore, b

(1) From the meaning of the title, it can be seen that f1 k, f2 qe is f1 f2, so qe k, e k, the direction of the uniform electric field is along the db direction.

2) The tentative charge is placed at point C:

eb＝e2＋e＝2e＝2k

So fb qeb 2k in the direction of db.

The value of CE is objective, and the value of electric potential is related to the selection of zero electric potential point, so it is impossible to have a positive answer to the above questions. In general, it is specified that the point of zero electric potential at infinity is specified, what about the above problem? Analysis with the example of a positive charge electric field (1):

The electric field line at point A is denser than that at point B, and the electric potential gradually decreases along the direction of the electric field line, so the electric field line must be higher at the largest part of E. The electric field of a negative-point charge is used as an example for analysis. As can be seen from the figure:

EC ed On the contrary, the student analyzed from the above two examples: the high E is not necessarily large. In summary, the projection conclusion E is not necessarily high at the big place, and the E at the height is not necessarily large.

The analysis of (2) can be from the equivalent of the same species, different point charge analysis of the same amount of the same kind of point charge line midpoint: according to the superposition of the field strength, this point e=0, and the selection of an electric field line infinitely close to this point can be seen: the electric potential decreases along the direction of the electric field line, to infinity is 0, then the point 0

The midpoint of the charge connection line of the equal heterogeneous point: From the figure, the perpendicular line of the charge connection of the two points is an equipotential surface and extends to infinity, this point = 0, and from the figure, this point e≠0Summary:

Where e is zero, it is not necessarily zero, and where it is zero, e is not necessarily zero. Conclusion: There is no direct relationship between field strength and electric potential.