3. Point charge field strength formula E=kQ/r² is point charge, vacuum, and static.
4. The electric field is a vector, which satisfies the law of vector superposition.
5. The field strength generated by any charged body can be regarded as a superposition of the field strength generated by dividing the charged body into a point charge.
6. Field strength is proportional to the power.
Common methods:
-: Compensation method
Example: It is known that the electric field generated by a uniformly charged sphere on the outside of the sphere is the same as the electric field generated by a charge at the center of the sphere and the amount of charge is equal. As shown in the figure,
radius R is uniformly distributed on a sphere with a charge of Q. There are two points A and B on the straight line through the center of the sphere O. The distance between O and B, B and A is R. Now, a spherical cavity is dug in the ball with OB as the diameter. If the electrostatic force constant is k, the volume formula of the ball is V=4πr³/3, then the magnitude of the field strength at point A is (A)
Example: The electric field generated by a uniformly charged spherical shell in the outer space of the ball is equivalent to the electric field generated by the charge concentrated at the center of the ball. As shown in the figure,
uniformly distributes positive charges on the hemispherical surface AB. The total charge is q, the spherical radius is R, and CD is the axis through the hemispherical vertex and the sphere center O. There are two points M and N on the axis. OM=ON=2R. It is known that the field strength of point M is E, so the field strength of point N is?
[Analysis]
is filled into a complete ball, and then combined to utilize symmetry.
As shown in the figure,
is added to the right hemispherical shell with charge amount +q and a right hemispherical shell with charge amount -q on the right side of the diameter A0B. This is equivalent to only the original "left half +q spherical shell". At this time, the field strength of point N can be regarded as a combination of the complete +2q spherical shell and the "right half -q hemispherical shell". From the symmetry, it can be seen that the field strength generated by the "right half-q hemispherical shell" at point N is equal to the field strength generated by the "left half-q spherical shell" at point M.
[Answer]
kq/2R²-E
Example: A uniformly negatively charged hemispherical shell, the center of the sphere is point O, AB is its axis of symmetry, plane L perpendicular AB divides the hemispherical shell into two, L and AB intersect at point M, and the N and M points on the axis of symmetry AB are symmetric about point O. It is known that the electric field strength at any point inside a uniformly charged spherical shell is zero. Take the potential at the infinity to be zero, and the potential at the distance from the point charge q is r as φ=kq/r. Assume that the electric field intensity of the left part of the plane L at point M is E₁, and the electric potential is φ₁; the electric field intensity of the right part at point M is E₂, and the electric potential is φ₂; the electric field intensity of the entire hemispherical shell at point M is E₃, and the electric field intensity of the electric field intensity at point N is E₄. The correct statement among the following is (D)
A. If the surface areas of the left and right parts are equal, then E₁＞E₂, φ₁＞φ₂
B. If the surface areas of the left and right parts are equal, then E₁＜E₂, φ₁＜φ₂
C. Only when the surface areas of the left and right parts are equal can E₁＞E₂, E₃=E ₄
D. Regardless of whether the surface areas of the left and right parts are equal, there is always E₁＞E₂, E₃=E₄
Example: A spherical surface is uniformly carried a positive charge, and the electric field intensity in the ball is zero everywhere. As shown in the figure,
O is the center of the ball, and A and B are the diameters. Two points, OA=OB, now the sphere is divided into two parts perpendicular to AB, C is the point on the cross section, remove the left hemispheric surface, and the charges carried by the right hemispheric surface are still distributed evenly, then (A)
A. The electric potentials of points O and C are equal
B. The electric field intensity of point A is greater than point B
C. The potential from A to B is increased first and then decrease
0 D. The electric field strength gradually increases from A to B along the straight line
[Analysis]
C point field strength direction horizontal. This is because the field strength of points a and c offset (the internal field strength is zero), and the field strength of points b and c is synthesized horizontally about the vertical face, so the field strength vertical plane, the vertical plane is an equipotential plane, and the electric potentials of points O and C are equal.
The key to solving this problem is to grasp the symmetry and find out the relationship between the electric field generated by the charges on the two spheres. The field strength of the left hemisphere at point A is equal to the field strength generated by the missing right hemisphere at point B. The opposite direction is the key to solving the problem.
A. For a complete charged sphere, the field strengths of each point on the middle droop surface of its internal AB are zero. It can be seen that the field strengths of each point on the middle droop surface are largely reversed. Because the electric fields of the left and right hemispheres face symmetrical with respect to the middle droop surface, the field strength directions of each point on the middle droop surface are perpendicular to the middle droop surface, so the field strength directions of each point on the middle droop surface are perpendicular to the middle droop surface, so the field strength directions of each point on the middle droop surface are perpendicular to the middle droop surface, so the field strengths of each point on the left hemisphere are perpendicular to the middle droop surface, and after the left hemisphere face is removed, the field strengths of each point on the middle droop surface are perpendicular to the middle droop surface. Then, after the left hemisphere face is removed, the field strengths of each point on the middle droop surface are perpendicular to the middle droop surface are perpendicular to the middle droop surface. The straight and middle drooping surface, that is, the middle drooping surface is an equal potential surface, so the electric potentials of points O and C are equal, so A is correct;
B and D, fill the hemispherical shell in the question into a complete spherical shell, and the charge is uniform. Assume that the electric field intensity generated by the left and right hemispheres at point A is E₁ and E₂, respectively. From the question, the electric field intensity inside the uniformly charged spherical shell is zero everywhere, then we know that E₁=E₂. According to the symmetry, the electric field intensity generated by the left and right hemispheres at point B is E₂ and E₁=E₂.
In the electric field shown in the figure, the electric field intensity of A is E₁, the direction is to the left, and the electric field intensity of B is E₂, the direction is to the left, so the electric field intensity of point A is the same as the electric field intensity of point B, so the field intensity from A to B cannot gradually increase, so B and D are wrong; C. According to the principle of electric field superposition, it can be seen that the direction of the electric field line on the x-axis is to the left, and the electric potential along the direction of the electric field line decreases, so the electric potential of point B is higher than the electric potential of point A, so C is wrong;
Therefore, choose: A.
Example: As shown in the figure, the charge P with a charge amount of Q is 2r away from the uniformly charged circle plate. The charge to the perpendicular line of this point passes through the geometric center of the plate. A and B are the distance between the two points on the perpendicular line to the plate is r, and the electrostatic force constant is k. If the electric field intensity at point B is 0, then the electric field intensity at point A is ()
Example: The 1/4 rings in the following options are the same size, the amount of charge has been marked in the figure, and the charge is evenly distributed, and each 1/4 rings are insulated from each other. The largest electric field intensity at the coordinate origin O is (B)
utilizing the equivalent charge model and symmetry
Example: As shown in the figure, the field strength generated by the uniformly charged 3/4 spherical shell at the point is equivalent to the field strength generated by arc BC, the field strength direction generated by arc BC, and the field strength direction generated by midpoint M of arc at point O.
3: equivalent method
xOy plane is the surface of an infinite conductor, which is filled with space of z≤0, and the space of z>0 is vacuum. If the point charge with the charge amount q is placed at z=h on the z-axis, an induced charge will be generated on the xOy plane. The electric field at any point in space is excited by the point charge q and the induced charge on the surface of the conductor. It is known that the field strength of the conductor is zero everywhere during electrostatic equilibrium, so the field strength at z=h/2 on the z-axis is (k is the electrostatic force constant) ()
Example: After exploration, a student found that the electric field between the point charge and the infinitely large grounded metal plate (as shown in Figure A) is exactly the same as the electric field distribution between the equal amount of heterogeneous point charges (as shown in Figure B). The distance OA from the charge q to MN at the middle point of C is L, and AB is a diameter on a circle with charge q as the center and L as the radius. So what is the magnitude of the electric field intensity at the point B?
Example: As shown in the figure,
3 is uniformly charged three equal length insulating rods form an equilateral triangle ABC, and P is the center of the triangle. When the charge amounts of AB and AC rods are both +q and the charge amounts of BC rods are -2q, the P point field is strong and small.Now take away the BC rod, and the charge distribution of AB and AC rods remains unchanged. After taking away the BC rod, is the field strength of point P?
[Analysis] When the BC rod is not removed, the BC rod (-2q) can be regarded as composed of +q and -3q. Then the field strength of the point P can be regarded as being produced by three +q rods and one -3q rod. From the symmetry, the field strength generated by the three +q rods at point P is zero, so the field strength E of the point P is generated by a -3q rod. After taking away the BC stick, it can be regarded as putting a +q stick and a -q stick at the BC stick position. is equivalent to putting nothing . Similarly, the field strength of point P is generated by a -q stick, which is E/3.
field strength is proportional to the power.
Example: As shown in the figure,
uniformly charged ring is Q, the radius is R, the center of the circle is O, and P is a point perpendicular to the axis of symmetry of the ring plane, OP=L, try to find the field strength of point P.
As shown in Figure A, a uniformly charged circular plate with a radius R, the charged amount per unit area is δ, and the electric field strength of any point P (coordinate x) on its axis can be based on the superposition principle of Coulomb's law and the electric field strength of the electric field strength Find:
direction along the x-axis. Now consider an infinitely large uniformly charged flat plate with a unit area charge of δ₀, and dig out a circular plate with a radius of r from it, as shown in Figure B. Then the electric field intensity of any point Q (coordinate x) on the axis of the circular hole is (A)
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