High School Physics Ask for detailed answers about electric potential.

1.The electric potential decreases in the direction of the electric field lines. 2.

The equipotential surface is perpendicular to the electric field lines. a.Option you can try to draw an equipotential surface and then use 1 to obviously be the potential high of p.

Option B utilizes that the electric field between 3,ON is stronger than OM, and UOM is larger than UOM when D is the same. The C option can take advantage of the equipotential surface drawn in A, where the potential at point O is smaller than that at point Q, w = q, and since point q is positive, the potential at point Q is greater than point O. , UMP is negative and the charge is negative, so the product is positive, which is the electric field force.

Do the right work. w in c is the electric potential energy at a certain point.

It is equal to the electric potential of a point multiplied by the amount of charge. w in d is the work done, which is equal to the potential difference between two points.

Multiply by the amount of charge.

D should be chosen!It is conceivable that this set of electric field lines is generated by a positive point charge, then the point charge must be at the intersection point of this group of electric field lines, and this intersection point must be on the y-axis!!

Let the intersection point be point a, then am must be greater than ap, ao must be less than aq, so the electric potential at point m is lower than that at point p! The potential of the o point is higher than that of the q point! Neither A nor C is right!!

on=om, but the electric field strength between ons is greater than that between oms, so answer b is also incorrect!

Draw an arc with n as the center of the circle, pn, mn, qn as the radius, and each arc as an isopotential line. The electric potential decreases gradually along the direction of the electric field line, so the electric potential at point P is higher than that at point m, and A is wrong.

om=on, according to e=u d, it can be seen from the figure that the electric field lines of on are denser than om, so the electric field strength of on is greater than om, and when d is equal, the potential difference of on is larger than om, and b is wrong.

A positive charge moves from o to q, the electric field force and the displacement are in the same direction, the electric field force does positive work, and the electric potential energy decreases, so the electric potential energy at point o is greater than the point q, and c is wrong.

A negative charge moves from m to point p, and the electric field force and the displacement do positive work in the same direction as the electric field force, d pair.

The electric field force received by the object is f=eq, and the direction is the same as the electric field, because the direction of motion and the direction of the electric field are opposite, the electric field force does negative work, and the change of electric potential energy can be judged according to the work done by the electric field force; The change in kinetic energy can be found by combining external forces to do work using the kinetic energy theorem

Solution: a. Since the electric field force and the direction of motion of the object are opposite, the electric field force does negative work, that is, the work done by overcoming the electric field force w=eqs, so a is correct;

b. The electric field force does negative work, and the electric potential energy increases, so B is wrong;

c. The amount of electric potential energy increase and the work done to overcome the electric field force are equal, so the electric potential energy increases EQS, so C is correct;

d. The object decelerates the motion, and the external force does negative work, and the kinetic energy decreases, which is obtained by the kinetic energy theorem: ekf combination. s=mas=, so d is correct

Therefore, ACD is chosen

At this time, the velocity of the particle is 0, and the kinetic energy is all used to overcome the electric potential energy of the electric field, so the electric potential energy is the kinetic energy of the particle.

Choose D, the horizontal direction does not consider gravity.

Because the electric field line is directed from the high potential point to the low electric potential point, and because the force direction of the negatively charged electron is opposite to the direction of the electric field line, the process of moving the electron from the high potential point to the low electric potential point is to overcome the electric field force to do work.

Regardless of whether it is a positive or negative charge, whenever work is done by overcoming the electric field force, the electric potential energy increases.

Because the direction of motion is opposite to the direction of the electric field force, the electric field force does negative work and the electric potential energy increases.

Electrons are negatively charged, and negative work is done along the electric field lines.

If the charged properties of the two point charges are opposite, then the direction of the electric field force does not change regardless of whether the two point charges are placed on Ab or Cd, and it is impossible for the particle to reciprocate.

If both point charges are positively charged, the two point charges should be on ab; If both point charges are negatively charged, the two point charges should be on the CD. The direction of the restoring force is always opposite to the displacement. However, unlike the simple harmonic oscillator, the magnitude of the restoring force is not proportional to the magnitude of the displacement. As shown in Fig.

First of all, the two point charges cannot be opposite in nature, otherwise the particles without muzzle velocity positively charged will inevitably move to the negative point charge, and will not make a round-trip in the middle, analyze cd, if you want to make a round-trip motion, you must be forced, the direction points to the equilibrium position, that is, the o point, the analysis of its motion, is to accelerate to o first, reverse force through o, decelerate to b to stop, and then reverse acceleration to o, so that the cycle can do round-trip motion, the key is with the force, c is the repulsion force, d is the gravitational force, all meet the requirements, about the electric potential, Here the system is the conversion of kinetic energy and electric potential energy, which is conserved by energy, and the kinetic energy at ab is 0, so the potential energy must be the same, so CD is chosen

Typing is not easy.

Hope, thank you.