How to calculate the electric field force on negatively charged
Publish: 2021-05-21 20:18:31
1. The formula of the ring is e = kqx / (x ^ 2 + R ^ 2) ^ (3 / 2) x is the distance from the point to the center of the ring, l is the distance from the point DL to the point θ The angle between L and X is k = 1 / 4 πε The method is to find the electric field of a point DL on the ring at that point. The electric field of the vertical axis is offset by the symmetry principle, and the horizontal direction de = K λ dl/l^2*cos θ And then integral E
2. If you know the distance and electric quantity, use f = Q1 * Q2 * k / (R ^ 2). Then judge the direction according to the electric property. Or use statics knowledge according to conditions.
3. A. According to the trajectory of particles, if the direction of electric field force is vertical upward, the particles will be negatively charged. So a is wrong and B is correct.
C. because the electric field is a uniform electric field, the electric field force on the particles is constant and the particles move in a constant speed curve. So C and D are correct.
BCD is selected
C. because the electric field is a uniform electric field, the electric field force on the particles is constant and the particles move in a constant speed curve. So C and D are correct.
BCD is selected
4. F = QE
for uniform electric field, e = u / d
is not f = Qu / D?
for uniform electric field, e = u / d
is not f = Qu / D?
5. A
if the initial velocity is zero, it can move along the electric field line
if the initial velocity is perpendicular to the electric field line and the velocity is appropriate, it can move in a uniform circular motion
if the initial velocity is zero, it can move along the electric field line
if the initial velocity is perpendicular to the electric field line and the velocity is appropriate, it can move in a uniform circular motion
6. If the equipotential surface moves in a circle only under the action of electric field force, the equipotential surface is required to be circular. C equipotential surface is mbbell shaped, and the centripetal force must point to the center of the circle, and the negative repulsive force is outward; It's also wrong to make an analogy
7. Solution: 
solution: 1. Because the electric field line formed by a positive point charge is a straight line, when the direction of the initial velocity of the negatively charged particle is in a straight line with the direction of the electric field line, the particle can move on the electric field line, and when the direction of the initial velocity of the particle is perpendicular to the electric field line, The particle can move in a uniform circular motion on the equipotential surface. The negatively charged particle can move in a uniform circular motion on any equipotential surface in the electric field formed by a positively charged point charge (such as equipotential surfaces a, B, C shown by the dotted line in Figure 1)< When the direction of the initial velocity of the particle is perpendicular to the electric field line, the particle will make centrifugal motion under the action of the force opposite to the electric field line, so ② is wrong. < br > ③ as shown in Figure 2, the electric field formed by two equal amount of the same kind of punctual charges a and B (the author adds four linear electric field lines Ao, Bo, OP and OQ), When the direction of the initial velocity of the negatively charged particle is in a straight line with the electric field line, the particle can move on the electric field line. As shown in Figure 2, the particle can move in a straight line on the linear electric field lines am, BN, AB, Op, OQ, etc. because the electric field intensity changes, the particle moves in a variable speed straight line. Considering the spatial and symmetry of the electric field distribution, when the direction of the initial velocity of the particle is in a, B, Op, OQ, etc In the vertical plane of the line B and perpendicular to the line B, the resultant force of Coulomb force on the particle always points to point O. the particle will move in a uniform circular motion in the vertical plane of the line a and B with the midpoint o of the line a and B as the center. In fact, PQ is the projection of the vertical bisector of the line a and B on the paper surface. In this plane, the potential distribution is like that of the line a and B If an isolated positive point charge is fixed at the midpoint o of line B, its potential distribution is as the equipotential surface shown by the dotted line in Figure 2, so ③ is correct< (4) two electric fields with equal negative point charges exert outward force on the negative charge, which is similar to the electric field of negative point charge, so it is impossible for the negatively charged particles to make uniform circular motion under the action of deviating from the center of the circle
solution: 1. Because the electric field line formed by a positive point charge is a straight line, when the direction of the initial velocity of the negatively charged particle is in a straight line with the direction of the electric field line, the particle can move on the electric field line, and when the direction of the initial velocity of the particle is perpendicular to the electric field line, The particle can move in a uniform circular motion on the equipotential surface. The negatively charged particle can move in a uniform circular motion on any equipotential surface in the electric field formed by a positively charged point charge (such as equipotential surfaces a, B, C shown by the dotted line in Figure 1)< When the direction of the initial velocity of the particle is perpendicular to the electric field line, the particle will make centrifugal motion under the action of the force opposite to the electric field line, so ② is wrong. < br > ③ as shown in Figure 2, the electric field formed by two equal amount of the same kind of punctual charges a and B (the author adds four linear electric field lines Ao, Bo, OP and OQ), When the direction of the initial velocity of the negatively charged particle is in a straight line with the electric field line, the particle can move on the electric field line. As shown in Figure 2, the particle can move in a straight line on the linear electric field lines am, BN, AB, Op, OQ, etc. because the electric field intensity changes, the particle moves in a variable speed straight line. Considering the spatial and symmetry of the electric field distribution, when the direction of the initial velocity of the particle is in a, B, Op, OQ, etc In the vertical plane of the line B and perpendicular to the line B, the resultant force of Coulomb force on the particle always points to point O. the particle will move in a uniform circular motion in the vertical plane of the line a and B with the midpoint o of the line a and B as the center. In fact, PQ is the projection of the vertical bisector of the line a and B on the paper surface. In this plane, the potential distribution is like that of the line a and B If an isolated positive point charge is fixed at the midpoint o of line B, its potential distribution is as the equipotential surface shown by the dotted line in Figure 2, so ③ is correct< (4) two electric fields with equal negative point charges exert outward force on the negative charge, which is similar to the electric field of negative point charge, so it is impossible for the negatively charged particles to make uniform circular motion under the action of deviating from the center of the circle8. 
solution: if a negatively charged particle is only in the electric field of a point charge, and the electric field force is perpendicular to the direction of velocity, then it moves in a uniform circular motion, then the point charge must be positive, And the electric field force provides the centripetal force. < br > If two point charges are superposed together in the electric field, if two positive point charges and the electric quantity is equal, the centripetal force is provided by the resultant force of the electric field force of the two point charges, and the direction of the resultant force is perpendicular to the velocity, so the particle is the center of the circle at the center of the line between the two charges and moves in a uniform circular motion< Br > so: AC
9.
Absolutely not. If the electric field line is perpendicular to the equipotential surface, then it is impossible to move on both the electric field line and the equipotential surface
but only moving on the equipotential surface can be satisfied, as shown in the figure, the red line is the vertical line of the two positive point charges, and the negatively charged particle is on the vertical line. At this time, the velocity V is perpendicular to the plane where the black line and the red line are located (for example, the black line is a horizontal line on your computer screen, and the red line is a straight line perpendicular to the black line on the screen, At this time, the resultant force of the two charges on the particle is perpendicular to the velocity. According to the symmetry, this relationship can be maintained in space. As long as the height of the particle on the vertical line is appropriate, the particle can make a circular motion
10. < Table > < tbody > < tr > < td > < br > < br > A: when the direction of the initial velocity of the particle is perpendicular to the electric field line, the particle will do centrifugal motion e to the force opposite to the electric field line< Br > b: because the electric field line formed by a positively charged point charge is a straight line, when the direction of the initial velocity of the negatively charged particle is in a straight line with the direction of the electric field line, the particle can move on the electric field line, and when the direction of the initial velocity of the particle is perpendicular to the electric field line, The particle can move in a uniform circular motion on the equipotential surface. The negatively charged particle can move in a uniform circular motion on any equipotential surface (such as equipotential surfaces a, B, C shown by the dotted line in Figure 1) in the electric field formed by a positively charged point charge, so option B is correct. < br > C: the force of the electric field with two equally negative point charges on the negative charge is outward, It is similar to the electric field of negative point charge, so it is impossible for a negatively charged particle to move in a uniform circular motion under the action of deviation from the center of the circle, so C is wrong< Br > D: as shown in Figure 2, the electric field formed by two identical positive point charges a and B (the author added four linear electric field lines Ao, Bo, OP and OQ), when the direction of the initial velocity of the negatively charged particle is in line with the electric field line, the particle can move on the electric field line, as shown in Figure 2, the particle can move on the linear electric field lines am, BN, AB, OP and OQ, Considering the space and symmetry of the electric field distribution, when the direction of the initial velocity of the particle is in the vertical plane of the line a and B and perpendicular to the vertical line, the resultant force of the Coulomb force on the particle always points to point O, and the particle will move in the vertical plane of the line a and B in the form of two charges a and B In fact, PQ is the projection of the vertical bisector plane of the two charge lines on the paper surface. In this plane, the potential distribution is like fixing an isolated positive point charge at the midpoint o of the two charge lines a and B, The distribution of its potential is shown on the equipotential surface as the dotted line in Fig. 2. Therefore, option D is correct. < br > therefore, BD < / td > < / TR > < / tbody > < / Table >
is selected
< br > A: when the direction of the initial velocity of the particle is perpendicular to the electric field line, the particle will do centrifugal motion e to the force opposite to the electric field line< Br > b: because the electric field line formed by a positively charged point charge is a straight line, when the direction of the initial velocity of the negatively charged particle is in a straight line with the direction of the electric field line, the particle can move on the electric field line, and when the direction of the initial velocity of the particle is perpendicular to the electric field line, The particle can move in a uniform circular motion on the equipotential surface. The negatively charged particle can move in a uniform circular motion on any equipotential surface (such as equipotential surfaces a, B, C shown by the dotted line in Figure 1) in the electric field formed by a positively charged point charge, so option B is correct. < br > C: the force of the electric field with two equally negative point charges on the negative charge is outward, It is similar to the electric field of negative point charge, so it is impossible for a negatively charged particle to move in a uniform circular motion under the action of deviation from the center of the circle, so C is wrong< Br > D: as shown in Figure 2, the electric field formed by two identical positive point charges a and B (the author added four linear electric field lines Ao, Bo, OP and OQ), when the direction of the initial velocity of the negatively charged particle is in line with the electric field line, the particle can move on the electric field line, as shown in Figure 2, the particle can move on the linear electric field lines am, BN, AB, OP and OQ, Considering the space and symmetry of the electric field distribution, when the direction of the initial velocity of the particle is in the vertical plane of the line a and B and perpendicular to the vertical line, the resultant force of the Coulomb force on the particle always points to point O, and the particle will move in the vertical plane of the line a and B in the form of two charges a and B In fact, PQ is the projection of the vertical bisector plane of the two charge lines on the paper surface. In this plane, the potential distribution is like fixing an isolated positive point charge at the midpoint o of the two charge lines a and B, The distribution of its potential is shown on the equipotential surface as the dotted line in Fig. 2. Therefore, option D is correct. < br > therefore, BD < / td > < / TR > < / tbody > < / Table > is selected
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