[Title] As shown in Figure A, Q₁ and Q₂ are two fixed point charges, and there are two points a and b on the extension line of their connection. The test charge in a positively charged area is only under the action of electric field force, and moves from point a to move along the line ab from point a to distance. When passing point b, the velocity is vb. The v-t image is shown in Figure B, which stipulates that the infinity is zero potential point. The following statement is correct (C)
A.Q₁ is negatively charged, Q₂ is positively charged
B. The potential of point b is zero
C. The field strength of point b is zero
D. Test the charge from point a to infinity, and the electric field force does the negative work
[Analysis]
From the v-t diagram, we can see that the ab section tests the charge to decelerate motion, and the field strength goes to the left; then b does the acceleration motion, and the field strength goes to the right. The acceleration a=0 at point b is the electrostatic force is zero, and the field strength is zero, that is, the electric fields excited by Q₁ and Q₂ cancel each other out, and the charge properties of the two points are opposite. Draw the field strength near point b as shown in the figure.
For option A, the answer is very confusing and I feel that the explanation is not strong.
Set point O as the origin and establish a one-dimensional coordinate system
From this ratio, we can see that on the right side of point b, the electric field intensity of excitation is greater than the electric field intensity of Q₂ excitation; on the left side of point b, the electric field intensity of Q₂ excitation is greater than the electric field intensity of Q₁ excitation.
can also use derivatives to calculate the rate of change of electric field intensity along the x-axis.
can also make the same conclusion using mathematical limits.
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