You have actually two dues on one axis. One charge of 

*
 is located at the origin, and also the other charge of 
*
 is situated at 4m. At what suggest along the axis is the electric field zero?

*
*


*


*



*


Correct answer:


Explanation:

The equation for an electric field native a point charge is

*

To discover the allude where the electric field is 0, we set the equations because that both charges equal to every other, due to the fact that that"s where they"ll publication each other out. Permit

*
 be the point"s location. The radius because that the very first charge would be
*
, and the radius because that the 2nd would it is in
*
.

You are watching: At what points along the x axis is the electric field zero.

*

*

*

*

*

*

*

Therefore, the only point where the electric field is zero is at

*
, or 1.34m.


Explanation:

To uncover where the electrical field is 0, we take the electrical field for each point fee and collection them equal to every other, due to the fact that that"s as soon as they"ll release each various other out.

*

The

*
"s can cancel out. 

*

*

*

*

*

*

*

Therefore, the electric field is 0 at

*
.


Imagine two suggest charges 2m far from each various other in a vacuum. Among the charges has actually a toughness of 

*
. If the force in between the particles is 0.0405N, what is the toughness of the 2nd charge?

*


Explanation:

The equation for pressure experienced by two suggest charges is

*

We"re do the efforts to uncover

*
, so us rearrange the equation to settle for it.

*

*

*

Now, we deserve to plug in our numbers.

*

*

Therefore, the toughness of the 2nd charge is 

*
.


Explanation:

The force in between two point charges is shown in the formula below:

*
, where 
*
 and 
*
 are the magnitudes that the suggest charges, 
*
 is the distance in between them, and 
*
 is a continuous in this case equal to 
*

Plugging in the numbers right into this equation provides us

*

*

*


Suppose there is a framework containing an electrical field the lies level on a table, as is shown. A positively charged fragment with charge 

*
 and mass 
*
 is shot v an early velocity 
*
 at an angle 
*
 to the horizontal. If this particle starts its journey at the negative terminal that a continuous electric field 
*
, which of the complying with gives one expression that signifies the horizontal street this bit travels while within the electrical field?


Explanation:

We are given a instance in i m sorry we have actually a frame containing an electric field lying flat on that is side. In this frame, a positively charged fragment is traveling through an electric field that is oriented such the the positively fee terminal is top top the opposite side of wherein the fragment starts from. We are being inquiry to find the horizontal street that this particle will take trip while in the electric field. Due to the fact that this structure is lie on that side, the orientation of the electric field is perpendicular come gravity. Therefore, the only force we need problem ourselves through in this case is the electrical force - we have the right to neglect gravity. However, it"s advantageous if we think about the positive y-direction as going in the direction of the optimistic terminal, and also the negative y-direction as going towards the an adverse terminal. It"s additionally important to establish that any kind of acceleration the is arising only happens in the y-direction. The is to say, there is no acceleration in the x-direction. We"ll begin by using the following equation:

*

We"ll need to discover the x-component that velocity.

*

*

Our next challenge is to uncover an expression for the time variable. To perform this, we"ll require to take into consideration the movement of the particle in the y-direction. Also, since the acceleration in the y-direction is constant (due to a consistent electric field), we have the right to utilize the kinematic equations.

*

And since the displacement in the y-direction won"t change, we can collection it equal to zero.

*

*

Just as we did because that the x-direction, we"ll need to take into consideration the y-component velocity.

*

We also need to uncover an different expression because that the acceleration term. We deserve to do this by noting the the electric force is providing the acceleration.

*

*

Also, it"s essential to mental our authorize conventions. Because the electrical field is pointing native the positive terminal (positive y-direction) to the an adverse terminal (which we characterized as the an adverse y-direction) the electric field is negative.

*

Now, plug this expression into the above kinematic equation.

*

*

Rearrange and fix for time.

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*

Now the we"ve uncovered an expression because that time, we can at critical plug this value into our expression for horizontal distance.