Physics, 26.12.2019 20:31 pgld7555

Find the electric field at a point in the z axis for a charge distribution of 4 point charges of magnitude q located at (d,0,0), (-d,0,0), (0,-d,0) and (0,d,0). (the coordinates are expressed in a cartesian coordinate system.)

Answers: 2

Show answers

Answers

Answer from: Drew2844

$E=4Qz/(4\pi e_{0} (d^{2}+z^{2})^{3/2})$ in the z direction is the electric field at a point in the z axis for a given charge distribution.

Explanation:

Even before we begin the actual calculation we have to note the symmetry of the charge distribution. If the charge on the z axis is held fixed then we note that, for every charge on either x or y axis there is an equal charge present in the negative part of the axis same distance away from the center. If we draw out the geometry of the situation we will see that the x component of the field on the charge on the z axis from the charge on the positive x-axis will cancel the x component of the field from the negative x-axis, so we will be left with just the z component of the field, since the force on the charge on the z axis is repulsive and upwards. We also see that the y component of the field on the charge on the z axis from the charge on the positive y-axis will cancel the y component of the field from the charge from the negative y-axis as well, so again we will be left with just the z component of the field pointing upward since the force on the charge on the z axis is repulsive and upwards. We can write down the field on the charge explicitly by adding the vector electric field contributions from the 4 charges(Using electric field formula derived from columbs law for each charge given below),

$E=Q/4\pi e_{0}r^2$

Just the magnitude. if you multiply this by the unit vector in the direction of the electric field, we get the electric field vector. We see the cancellation happen mathematically like below,

$E = Q*(-di+zk)/(4\pi e_{0}(d^2 + z^2)^{3/2} ) +Q*(di+zk)/(4\pi e_{0}(d^2 + z^2)^{3/2} ) +Q*(-dj+zk)/(4\pi e_{0}(d^2 + z^2)^{3/2} ) +Q*(dj+zk)/(4\pi e_{0}(d^2 + z^2)^{3/2} )$

The magnitude of the electric field is the same from all four charges as shown by the 4 terms above, only their direction changes as shown. If we just simplify by adding the components we obtain,

$E=4Qz.k/(4\pi e_{0} (d^{2}+z^{2})^{3/2})$

The electric field. The direction of the field is upward(since the electric field vector contains only k the unit vector in the z direction.) because of the reasoning above and the mathematical simplification in the second last equation whereby the x components and the y components of the field at the point z on the z-axis cancel out due to the symmetrical distribution of the charges.

Answer from: Quest

answer: (431,2)

step-by-step explanation:

Adog's heart beats 431 times in 2 minutes. about how many times does a dog's heart beat in 1 minute?

Answer from: Quest

i think it would be 5

Another question on Physics

Physics, 21.06.2019 20:30

Water at room temperature of 20.0°c is poured into an aluminum cylinder which has graduation markings etched on the inside. the reading in the graduations is 300.0 cc. the cylinder with the water in it is then immersed in a constant temperature bath at a temperature of 100°c. what is the reading for the level of water on the graduations of the cylinder after the water and the cylinder reach thermal equilibrium with the bath? the volume coefficient of expansion of water is 2.07 x 10^-4 k-1, and the linear coefficient of expansion of aluminum is 23.0 x 10^-6 k-1. a) 305.0 cc b) 304.0 cc c) 303.5 cc d) 303.3 cc e) 304.5 cc

Answers: 3

Answer

Physics, 22.06.2019 09:00

Atelevision set that has been running for a while heats up even the air around it. which two laws of thermodynamics could be used to analyze this scenario

Answers: 1

Answer

Physics, 22.06.2019 15:00

When is a current produced? when the terminals of an electrochemical cell are connected by a wire if the electric circuit is opened in an electrochemical cell if the electrolyte is removed from an electrochemical cell when the electrodes are reversed in an electrochemical cell

Answers: 3

Answer

Physics, 22.06.2019 19:40

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (for example, would the velocity of a neutral particle be selected by passage through this device? ) the explanation of this is that the mass and the charge control the resolution of the device--particles with the wrong velocity will be accelerated away from the straight line and will not pass through the exit slit. if the acceleration depends strongly on the velocity, then particles with just slightly wrong velocities will feel a substantial transverse acceleration and will not exit the selector. because the acc

Answers: 1

Answer

You know the right answer?

Find the electric field at a point in the z axis for a charge distribution of 4 point charges of mag...

Questions