Fixed two dissimilar point charges Q1 and Q2 with a charge of 1 4 with a distance of d, 100

On the line connecting Q1 and Q2 (you may wish to set Q1 to the left of Q2), take the point where Q1 is located as the origin O, and let the coordinates of Q3 be X, then:

The electric field generated by q1 at q3 is: kq1 x

The electric field generated by q2 at q3 is: kq2 (d-x) Regardless of the electrical properties and electricity of q3, as long as the combined field strength of q1 and q2 at q3 is 0, q3 is not under force and is in equilibrium, so the equilibrium position x of q3 satisfies the following equation:

kq1 x +kq2 (d-x) =0 (q3 is on the left side of q1) to obtain: x=d =-d

kq1 x -kq2 (d-x) =0 (q3 is on the right side of q1) This equation has no solution.

Therefore, when Q3 is at D to the left of Q1, it is in a state of force equilibrium.

It has the same electrical properties as Q1.

The position is like this: Q3 Q1 Q2 is always on the side with the lower charge, not in the middle and on the side with the larger charge.

There is no requirement for the amount of electricity.

Position l=d ( (q2 q1)-1).

Let the slave mark of Q1 be (0,0), let the coordinates of Q2 be (0+d,0), and the coordinates of Q3 be (D+X,0).

To achieve equilibrium between the three points, q3 and q1 q2 are in the same straight line.

Suppose q3 has the same charge as q1.

Q3 is subjected to the force of Q1 = kq1q r1 2 to the right in the direction.

Q3 is subjected to the force of Q2 = kq2q R2 2 to the left in the direction.

The magnitude of the above two forces is equal, and the direction is opposite q3 in order to be in equilibrium.

So: k*q1*q3 (d+x) 2=k*q2*q3 x 2 Solve the equation to get x1=-2d 3 x2=-2d

1) If x=-2d 3 and q3 is between q1 and q2, then the coordinates of q3 are (d 3,0).

Q2 is subjected to the force of Q1 = K*Q1*Q2 D 2Q2 is subjected to the force of Q3 = K*Q3*Q2 (2d 3) 2So: k*q1*q2 d 2=k*q2*q3 (2d 3) 2 Solve the equation to obtain: q3=4q1 9

2) If x=-2d and q3 is to the left of q1, then the coordinates of q3 are (-d,0)q2 is subjected to the force of q1 = k*q1*q2 d 2q2 is subjected to the force of q3 = k*q3*q2 (2d) 2So: k*q1*q2 d 2=k*q2*q3 (2d) 2 Solve the equation to obtain: q3=4q1

2.Suppose Q3 has a dissimilar charge with Q1.

Follow the method above.

I'll show you ...

This q3 can be positive or negative, but it must be on the line of two charges, and it can be between the two, outside of the two, as long as the charge can be balanced, f=kqq d2, 2 is squared! For example, just assume a situation: q1+, q2-, q3+ can be arranged in q3, q1, q2, and so on, taking into account the repulsion of q1, and the gravitational force of q2 makes the two forces the same magnitude.

In other cases, you have to push it yourself, the same reason.

Sold by e=kq trillions of 2.

kq1/r^2=kq2/(d-r)2

r=√q1*d/(√q1+√q2)

The distance between the Lie cavity and q1 is q1*d ( Q1+ Q2) where the field strength of the two-point charge connection line is zero

I'll give you the solution idea:

This question is actually a test of Newton's first law.

If the charge q is stationary between ab, the force must be balanced i.e. Coulomb force from q1 = Coulomb force from q2 (equal in magnitude and opposite in direction) because q1 and q2 are both positive charges, q can be positive or negative q q must be negative if all three charges are at rest.

According to Newton's first law, no matter which charge is forced, it is balanced against q1, there is.

Gravitational force from q = repulsive force from q2.

There is it to Q. Gravitational force from q1 = gravitational force from q2.

For Q2, there is.

Gravitational force from q = repulsive force from q1.

Knowing the truth, the rest is the math problem.

The big clip is small, and the same clip is different.

So be sure to put a negative charge on the left side of A.

Then use the force equilibrium equation to solve two equations. The formula is not easy to play.

The formula is Coulomb's law.

First of all, the ordinate is ep. So m is on the x-axis. The electric potential is definitely 0. Then the electric potential rises to the left and right at point n. You know that the electric potential should be lowered along the electric field lines. So there must be no electric field lines at this point. So the field strength must be 0

Then to the left of the point m, the potential is greater than 0The electric potential around the positive charge is greater than 0. So Q2 is positively charged.

And the electric potential on the right side of the m point is less than 0So Q1 must have a negative point, and it must have a larger charge than Q2. Because the distance between the dots to the right of point m is obvious, q2 is significantly closer than q1.

In this way, only Q1 has a large battery. It has a greater impact on the right side of the m-point.

It can be seen from the figure that the potential energy decreases and then increases during the process of the tentative charge from a sufficient distance to the o point, and the point n is the demarcation point for the decrease and increase of the electric potential energy, which indicates that the force on the tentative charge before and after the n point is the gravitational force and then the repulsion force, and the combined field strength of the n point is zero.

The M point is the demarcation point where the electric potential energy changes from negative to positive, and the electric potential energy is zero, so the M potential is zero and the electric potential between Mo is positive, so Q2 is positively charged; The combined field strength of the M point is zero kq1 r12=kq2 r22, and r1 > r2, so q1 > q2

e=k*q/r^2

The absolute value of q is the same, in the line where q1 and q2 are located, if they are outside the two charges, r must be unequal, and e1 is not equal to e2

If there are two points between charges, then at the midpoint of the line Q1 and Q2, there is only one point of e1=e2.

Hope you help!

The formula for the strength of the telecommunication field of the point charge: e=kq r 2

e1=e2 then: kq1 r1 2=kq2 r2 2——2q2 r1 2=q2 r2 .

Then the point of e1=e2 has two points:

Point A is connected between Q1Q2 and Q2, and the distance from Q2 is R, and the distance from Q1 is 2*R, R+ 2*R=L, R=L (1+ 2).

e=e1+e2=k2q2/(√2*r)^2+kq2/r^2=2*kq2/r^2=2*kq2/(l/(1+√2))^2=(6+4√2)*kq2/l^2；

Point B is close to the outside of Q2 on the line of Q1Q2, and is away from Q2(2-1)*L, e=E1-E2=0

Summary. Kiss <>

Hello, I come to charge Q1 = 2 * 10 -6C and Q2 = 4 * 10 -6C are 10cm apart, find the position where the electric field strength on the line of the two charges is zero is the electric field strength between the charges can be calculated by the electromotive force formula: E = K * Q1 r 2 where e is the electric field strength, k is a constant (equal to x 10 9 n*m 2 c 2), Q1 is the amount of charge of charge 1, and r is the distance from charge 1 to charge 2.

The distance between Q1=2*10 -6C and Q2=4*10 -6C is 10 cm, and the position where the electric field strength on the connection line of the two charges is zero.

Okay. Relatives Shisui early [Smile Jane] <>

Hello, I come to charge Q1 = 2 * 10 -6C and Q2 = 4 * 10 -6C are 10cm apart, find the electric field strength on the line of the two charges is the position where the electric field strength is zero is the electric field strength between the charges can be calculated by the electromotive force formula: e = k * q1 r 2 where e is the electric field strength, k is a constant (equal to x 10 9 n*m 2 c 2), q1 is the amount of charge of charge 1, r is the distance from charge 1 to charge 2.

Good drops, you go on.

Kiss <>

For the position where the electric field strength between two charges is zero, we can set its distance x, so the absolute equation that I want to solve is: k * q1 x 2 = k * q2 10cm-x) 2 simplified: x = 10cm * sqrt(q1 q2) According to the data given in the question, we can calculate the value of x as:

x = 10cm * sqrt(210 -6c 410 -6c) = 5cm, so that the electric field strength on the line of the two charges is zero at a distance of 5 cm between the two charges.

Does this mean that both stores are positively charged by default?

Dear, yes.

What does that sqrt mean?

Dear, that's the meaning of open square root, because it can't be displayed on the keyboard and can only be replaced in English.

1. C is subjected to the repulsive force given by A and the attraction given by B, and the resultant force of the two forces is the electrostatic force sought.

The magnitude of the repulsive force given by A to C is fac k*q 2

The magnitude of attraction given by r 2b to c is fbc k*q 2r 2 and the angle between the directions of these two forces is 120 degrees, and their resultant force is f k*q 2r 2 , and the direction is 60 degrees with the two forces permeable slippery lead.

2. A is subjected to the repulsive force given by C and the attraction given by B, and the resultant force of the two forces is the electrostatic force sought.

The magnitude of the repulsive force given by c to a is fca k*q 2

The force of attraction given by r 2b to a makes the spike size fba k*q 2r 2r 2 good at the angle between the two directions of force is 120 degrees, and their resultant force is f k*q 2

r 2 , the direction is at an angle of 60 degrees to those two forces.

The two positive charges of q2 are placed separately.

Point A. and point b, the two points are separated by l on a smooth insulated semicircular with l diameter, wearing a charged ball + q (regarded as a point charge), and equilibrium at point p regardless of the gravity of the ball is known, then, the relationship between the angle between pa and ab and q1 and q2 should satisfy ( ).