What Is the Net Electric Field at the Midpoint Between Two Charged Particles?

In summary, the problem involves two particles with charges of -5.00 microCoulombs and +5.00 microCoulombs, respectively, fixed in an x-y coordinate system at specific points. To find the net electric field at the midpoint between the particles, one must use the equation 1/4piEo x q/r^2, where q is the charge of the particle and r is the distance from the particle to the midpoint. The field vector must point along the line connecting the two particles, so one must determine the unit vector along that line. The resulting net electric field is not simply the sum of the electric fields from each particle, as the direction of the fields must also be taken into account. Further
  • #1
threewingedfury
29
0
Two particles are fixed to an x-y coordinate system:
particle 1 of charge -5.00 microCoulombs lies on the x-axis at x = + 6.00 cm and particle 2 of charge +5.00 microCoulombs lies on the y-axis at y = + 8.00 cm. Midway between the particles, what is their net electric field, in unit vector notation?

So I got:
1/4piEo x q/r^2 - 1/4piEo x q/r^2

(1/(4pi(8.85x10^-12)) x (-5 x 10^-6)/(.06)^2)i - (1/(4pi(8.85x10^-12)) x (5 x 10^-6)/(.08)^2)j

-1.248e7i + 7024847j

Is this anywhere near right?
 
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  • #2
Not quite.

The electric field vector must point along the line from the positive charge to the negative charge, so one needs to determine the unit vector along that line.

The problem asks for the field at the midpoint between the charges, and so one needs the distance between the positive charge and the midpoint, and the negative charge and the midpoint. For one point at (x, 0) and the other at (0, y), think of (x/2, y/2).
 
  • #3
so can anyone give me a better idea?
 
Last edited:
  • #4
pretty please?
im a bit confused
 
  • #5
So is it more like:

(1/(4*π*ε_0)*(5.00 μC / (3.00 cm)^2) for particle 1 and give it an 'i' and -1/(4*π*ε_0)*(5.00 μC / (4.00 cm)^2) for particle 2 and give it a j?

Since that's r/2?
 

What is an electric field of a plane?

The electric field of a plane is a measure of the strength and direction of the electric force at any point in space around a flat, two-dimensional surface. It is a vector quantity, meaning it has both magnitude and direction.

How is the electric field of a plane calculated?

The electric field of a plane can be calculated using the formula E = σ/ε0, where E is the electric field, σ is the surface charge density, and ε0 is the permittivity of free space. This formula assumes that the plane has a uniform charge distribution.

What is the direction of the electric field of a plane?

The direction of the electric field of a plane is perpendicular to the surface of the plane. This means that the electric field lines are parallel to the plane, pointing away from positively charged planes and towards negatively charged planes.

How does the distance from the plane affect the electric field?

The electric field of a plane decreases with distance from the plane. As you move farther away from the plane, the electric field becomes weaker. This is because the electric field is inversely proportional to the distance squared.

What are some real-life applications of the electric field of a plane?

The electric field of a plane has many practical applications, such as in the design of capacitors, where two parallel plates act as planes to store electric charge. It is also important in understanding the behavior of lightning strikes and the use of electric fields in electrostatic precipitators to remove particles from air. Additionally, the electric field of a plane is used in the production and control of plasma in fusion reactors.

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