You have actually two charges on an axis. One charge of 

*
 is located at the origin, and the other charge of 
*
 is situated at 4m. In ~ what point along the axis is the electric field zero?

*
*


*


*



*


Correct answer:


Explanation:

The equation because that an electrical field native a point charge is

*

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

*
 be the point"s location. The radius because that the first charge would be
*
, and also the radius for the second would it is in
*
.

You are watching: Find a point where the electric field is zero.

*

*

*

*

*

*

*

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

*
, or 1.34m.


Explanation:

To discover where the electric field is 0, us take the electrical field for each point fee and collection them equal to each other, because that"s once they"ll cancel each other out.

*

The

*
"s can cancel out. 

*

*

*

*

*

*

*

Therefore, the electric field is 0 at

*
.


Imagine two suggest charges 2m far from each other in a vacuum. One of the charges has a stamin of 

*
. If the force 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 make the efforts to find

*
, so we rearrange the equation to fix for it.

*

*

*

Now, we can plug in ours numbers.

*

*

Therefore, the strength of the 2nd charge is 

*
.


Explanation:

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

*
, where 
*
 and 
*
 are the magnitudes of the point charges, 
*
 is the distance in between them, and 
*
 is a constant in this case equal to 
*

Plugging in the numbers right into this equation provides us

*

*

*


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

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


Explanation:

We are offered a situation in which we have a structure containing an electrical field lying flat on that is side. In this frame, a positively charged fragment is traveling v an electric field the is oriented such the the positively charged terminal is on the opposite side of whereby the particle starts from. We are being inquiry to discover the horizontal distance that this bit will travel while in the electric field. Since this frame is lied on the side, the orientation that the electrical field is perpendicular to gravity. Therefore, the only force we need problem ourselves v in this case is the electric force - we have the right to neglect gravity. However, it"s advantageous if we think about the optimistic y-direction as going in the direction of the confident terminal, and the an adverse y-direction as going in the direction of the an adverse terminal. It"s additionally important to establish that any type of acceleration the is arising only wake up in the y-direction. That is to say, over there is no acceleration in the x-direction. We"ll start by using the adhering to equation:

*

We"ll need to find the x-component of velocity.

*

*

Our next an obstacle is to uncover an expression for the moment variable. To execute this, we"ll require to consider the motion of the bit in the y-direction. Also, because the acceleration in the y-direction is constant (due to a consistent electric field), we have the right to utilize the kinematic equations.

*

And due to the fact that the displacement in the y-direction won"t change, us can collection it same to zero.

*

*

Just together we did for the x-direction, we"ll require to take into consideration the y-component velocity.

*

We additionally need to find an alternate expression because that the acceleration term. We can do this by noting the the electric force is providing the acceleration.

*

*

Also, it"s vital to remember our authorize conventions. Due to the fact that the electrical field is pointing from the hopeful terminal (positive y-direction) to the an unfavorable terminal (which we identified as the an unfavorable y-direction) the electrical field is negative.

*

Now, plug this expression right into the above kinematic equation.

*

*

Rearrange and solve for time.

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*

Now the we"ve found an expression for time, we deserve to at last plug this value right into our expression because that horizontal distance.