[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)
![[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 positively charged test charge in one area is only under the action of electric field force, and moves from point a to move along t - DayDayNews](https://cdn-dd.lujuba.top/img/loading.gif)
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
![[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 positively charged test charge in one area is only under the action of electric field force, and moves from point a to move along t - DayDayNews](https://cdn-dd.lujuba.top/img/loading.gif)
[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.
![[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 positively charged test charge in one area is only under the action of electric field force, and moves from point a to move along t - DayDayNews](https://cdn-dd.lujuba.top/img/loading.gif)
For option A, the answer is very confusing and I feel that the explanation is not strong.
![[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 positively charged test charge in one area is only under the action of electric field force, and moves from point a to move along t - DayDayNews](https://cdn-dd.lujuba.top/img/loading.gif)
![[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 positively charged test charge in one area is only under the action of electric field force, and moves from point a to move along t - DayDayNews](https://cdn-dd.lujuba.top/img/loading.gif)
Set point O as the origin and establish a one-dimensional coordinate system
![[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 positively charged test charge in one area is only under the action of electric field force, and moves from point a to move along t - DayDayNews](https://cdn-dd.lujuba.top/img/loading.gif)
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.