Questions and answers about two point charges in vacuum

• There are two equal amounts of the same point charges in the vacuum, both of which are q, the distance between them is r, and what is the field strength at the midpoint of the two point charge line

  Fatal Frame. Equally large and inversely added to zero.

  physics

  1 answer

• High School Physics

  The answer upstairs is quite correct

  Learning help

  2 answers

• Point charge in vacuum

  There are two point charges in the vacuum, and the electrostatic force between them does not change F. If the amount of charge carried by each of them is kept unchanged, and the distance between them is reduced to half of the original, then the electrostatic force between them is equal to four times of the original, that is, 4F

  physics

  1 answer

• There are two point charges Q1=3 times 10 to the negative octave and Q2=- 3 times 10 to the negative octave in the vacuum. They are 0.1M apart. Find the field strength of point A in the electric field

  The field strength of point charge 1 at point A is E (1)=kQ/R2=2.7 times 10 to the third power. Similarly, the field strength of point charge 2 at point A is E (2)=kQ/R2=2.7 times 10 to the third power. Since the distance between the two point charges and point A is equal, it is an equilateral triangle, and E (1) and E (2) form an external angle of 120 degrees. Its vector sum is E=2.7 times 10 to the third power

  mathematics

  1 answer

• There are two point charges in the vacuum, 20cm apart, Q1=2 times 10 to the - 8 power, Q2=- 8 times 10 to the - 6 power. What is the electric field intensity at the point in the connection between the two point charges. (The process should be detailed.)

  E=KQ/R ^ 2 9 * 10 ^ 9/(0.1) ^ 2 * (2 times 10 to the - 8 power, 8 times 10 to the - 6 power)=7218000V/m

  mathematics

  1 answer

• In High Two Physics, there are two point charges A and B in the vacuum. The amount of charge is - Q and 2Q respectively. If they are separated by L, what is the field strength at the midpoint O of the two point charge line?

  Use Coulomb's law F=K (QAQB)/r ^ 2 to calculate the electric field force on the charge. The electric field force is a pair of interaction forces. From this, according to the size of their respective charges, calculate their respective field strength E from E=F/q, under vector composition! Is the field strength at the midpoint!

  physics

  1 answer

• There are two point charges in the vacuum. The point charges are respectively q1=5 * 10 ^ - 3C and q2=- 5 * 10 ^ - 2C. They are 15cm apart. How to quote the third point charge, positive

  Heterogeneous charges attract each other and homogeneous charges repel each other. The three charges are static, and the electrostatic force received is balanced. After analysis, the third charge is negative, and q1 is between q2 and q3. Let q3 be X at the distance from q1. For q1 formula: kq1 * q2/r ^ 2=kq1q3/x ^ 2 for q2 formula: kq1 * q2/r ^ 2=kq2q3/(r x) ^ 2, the values of q3 and X can be obtained by solving the equation. Satisfied, please adopt

  mathematics

  1 answer

• Please explain how to solve the physics problem of Senior One, thank you!

  [Original] Two identical clips and one different, two big clips and one small, my pithy formula. Judging from Q1 and Q2, the third charge, if the third charge is negative, should be placed at... (See the figure). Take Q3 and Q1 as the research object. Let the distance between Q1 and Q3 be X, and the three charges are in a straight line. Formula (K (Q2) (Q3))/((X+15) ^ 2)=(K (Q1) (Q3))/(X ^ 2) and (K (Q1) (Q3))/

  physics

  1 answer

• Now the third point charge is introduced, then what is the amount of charge it should carry, and where can it be placed to make the three point charges stay at rest

  Since the two points are visible and mutually attractive, an outward force should be applied to them at the same time. You can put a negative charge on the left side of q1 or a positive charge on the right side of q2 (the charge is substituted into the absolute value, regardless of the positive and negative sign). If it is at the left of q1, xcm, q1q2/r ^ 2=q1q3/x ^ 2 q1q2/r ^ 2=q2q3/(r x) ^ 2, the solution is x=r=15cm, q3=- 5 * 10 ^ (- 3) C

  physics

  1 answer

• Senior 2 Physics

  Heterogeneous charges attract each other and homogeneous charges repel each other. The three charges are static, and the electrostatic force received is balanced. After analysis, the third charge is negative, and q1 is between q2 and q3. Let the distance from q3 to q1 be X, For q1 formula: kq1 * q2/r ^ 2=kq1q3/x ^ 2 For q2 formula: kq1 * q2/r ^ 2=kq2q3/(r+x) ^ 2 The values of q3 and X can be obtained by solving the equation.

  C physics

  1 answer

• High II physical electric field 4

  F=kQ1Q2/r2 a=F/M At the beginning, the acceleration of A is a, so k.q1.q2/L ^ 2=ma After a period of time, the acceleration of b is a, and then the charge distance is set as L1 Then F=(2m). a=2. (ma)=2.k.q1.q2/L ^ 2=k.q1.q2/L1 ^ 2, so 2. L1 ^ 2=L ^ 2, so L1=√ 2/2L Because two particles are affected by

  physics

  1 answer

• The electric field problem of high two physics

  At this time, the distance between two charges is (1/√ 2) L, and the rate of point charge A is 2V

  physics

  1 answer

• Coulomb's Law

  Let q be r from Q1 and R from Q2 F=KQ1*q/r^2=KQ2*q/R^2， When K and q are removed from the above equation, Q1/r ^ 2=Q2 */R ^ 2 Q1>Q2 Natural r>R

  physics

  2 answers

• There are two point charges q1 and q2 in the vacuum, where q1 is positive, q2 is negative, q1=8q2, and the other point charge p distance q1 is r1 distance q2 is r2

  It can be seen from the superposition principle of electric field that: the electric field strength generated by two charges at point P should be calculated first, and then solved according to the vector synthesis rule. The electric field strength of point charge Q1 at point P is E1=k, and the direction is as shown in the figure; The field strength of Q2 at P is E2=k, and the direction is as shown in the figure. To make the direction of the combined field strength E parallel to the connecting line of two point charges, then △ PE1E is similar to △ Q1PQ2. According to the similar triangle principle, we can get:, which can be solved by:=2

  physics

  1 answer