# What will be the acceleration of an electron when it is at the origin?

• AxM=Fam
In summary, the problem involves two point charges (+3.00μC and -2.50nC) at different positions, and an electron with a known mass (9.109X10^-31 kg) at the origin. The electrostatic force formula is used to calculate the force on the electron from each point charge, and the resulting vectors are added to find the net force on the electron. The direction of the force vector will depend on the signs and positions of the charges. To find the acceleration of the electron, the net force is divided by the electron's mass. It is important to draw a diagram and sketch the force vectors in order to better understand the direction of the forces.
AxM=Fam
1. +3.00μC point charge is at x=1.00cm and y=.750cm. -2.50nC point charge is at x=0 and y=-5.00mm. What will be the acceleration of an electron (Me=9.109X10^-31 kg) when it is at the origin (only force acting on the electron is the Coulomb force)?

2. K=8.988X10^9 N(m^2/c^2)

Fc= K(qs qd/r^2sd)(direction rsd)
I believe this is the electrostatic force formula? s=source d=detector.

3. This is my attempt at the problem.
Fc=(K)(3.00X10^-6C)(2.50X10^-9C)/(2.5625X10^-4 m^2) [.624695i+.780869j]
The 2.50X10^-9C was a negative to begin with but I redistributed the negative to the i and j hat direction to make them positive.

|Fcsd|=.26306N
θ=Tan^-1(.780869/.624695)=51.3degrees

I cannot figure out how to get the acceleration of this electron?
Do I use F=ma
.26306N/(9.109X10^-31 kg) = 2.89X10^29 m/s^2

I am probably using the wrong formula to begin with can I please get some help with this.

AxM=Fam said:
1. +3.00μC point charge is at x=1.00cm and y=.750cm. -2.50nC point charge is at x=0 and y=-5.00mm. What will be the acceleration of an electron (Me=9.109X10^-31 kg) when it is at the origin (only force acting on the electron is the Coulomb force)?

2. K=8.988X10^9 N(m^2/c^2)

Fc= K(qs qd/r^2sd)(direction rsd)
I believe this is the electrostatic force formula? s=source d=detector.

3. This is my attempt at the problem.
Fc=(K)(3.00X10^-6C)(2.50X10^-9C)/(2.5625X10^-4 m^2) [.624695i+.780869j]
The 2.50X10^-9C was a negative to begin with but I redistributed the negative to the i and j hat direction to make them positive.

|Fcsd|=.26306N
θ=Tan^-1(.780869/.624695)=51.3degrees

I cannot figure out how to get the acceleration of this electron?
Do I use F=ma
.26306N/(9.109X10^-31 kg) = 2.89X10^29 m/s^2

I am probably using the wrong formula to begin with can I please get some help with this.
Hello AxM=Fam . Welcome to PF !

It looks to me like you're finding the force on one of the charges due to the other.

You need to find the force that each charge exerts on the electron, then add those forces vector-wise.

By the Way: 0.624695i + 0.780869j is not a unit vector.. It's magnitude is approximately 1.4 .

I think I am getting a better understanding of the problem. I was missing the charge for the electron which was e=-1.602 x 10^-19. I would just have to figure the net force of the two point charges acting on the electron. From there would I just have to divide the net force by the given electron mass?

I am having a hard time understanding the direction of the vector +3.00μC acting on the electron. Is + always attracted to the negative meaning it will go towards the negative or will the negative go up towards the +. This will help me when I begin to add both vectors to get the net force. I know that like signs will repel so the -2.50nC will push the electron up .5cm.

Can I get more help on this (direction of the +3.00μC on electron is confusing me)? Hope I am at least getting the right idea on how this works.

Opposite charges attract, like charges repel. A negative electron will be pulled towards a positive charge and repelled from a negative charge. Thus the direction of any resulting force vector will depend upon the signs of the interacting charges and their relative positions in the coordinate system. It's always a good idea to draw a diagram and sketch in an approximation of the force vectors.

I would first verify the accuracy of the given information and ensure that all necessary units are consistent. Assuming the given information is correct, the first step would be to calculate the distance between the two point charges using the given coordinates. The distance between the two charges is 2.5625X10^-4 m.

Next, I would use the electrostatic force formula to calculate the magnitude and direction of the force acting on the electron at the origin. As you correctly stated, the formula is Fc= K(qs qd/r^2sd)(direction rsd). In this case, the source charge (qs) is +3.00μC and the detector charge (qd) is -2.50nC. Plugging in the values, we get Fc= (8.988X10^9 N(m^2/c^2))(3.00X10^-6C)(-2.50X10^-9C)/(2.5625X10^-4 m^2) = -0.26306 N.

Note that the negative sign indicates that the force is attractive, pulling the electron towards the point charge at (x=1.00cm, y=0.750cm). The direction of the force can be calculated using vector addition, as you did in your attempt. The direction of the force vector is 51.3 degrees, as you correctly calculated.

To find the acceleration of the electron, we can use Newton's second law, F=ma. Since we already know the magnitude of the force, we can rearrange the equation to solve for the acceleration. a=F/m= (-0.26306 N)/(9.109X10^-31 kg) = -2.89X10^29 m/s^2.

Note that the negative sign indicates that the acceleration is in the opposite direction of the force vector, towards the positive charge at (x=1.00cm, y=0.750cm). This is a very large acceleration, which is expected given the small mass of an electron and the strong force acting on it.

In summary, the acceleration of the electron at the origin is -2.89X10^29 m/s^2, towards the +3.00μC point charge at (x=1.00cm, y=0.750cm).

## 1. What is the formula for calculating the acceleration of an electron at the origin?

The formula for calculating the acceleration of an electron at the origin is a = F/m, where a is the acceleration, F is the net force acting on the electron, and m is the mass of the electron.

## 2. How does the electric field affect the acceleration of an electron at the origin?

The electric field plays a significant role in determining the acceleration of an electron at the origin. The direction of the electric field determines the direction of the force acting on the electron, which in turn affects its acceleration.

## 3. Can the acceleration of an electron at the origin be negative?

Yes, the acceleration of an electron at the origin can be negative. This indicates that the electron is experiencing a force in the opposite direction of its motion.

## 4. Does the initial velocity of an electron at the origin affect its acceleration?

Yes, the initial velocity of an electron at the origin can affect its acceleration. If the electron has an initial velocity, it will have a change in velocity and therefore experience acceleration.

## 5. Can the acceleration of an electron at the origin change over time?

Yes, the acceleration of an electron at the origin can change over time. This can occur if the net force acting on the electron changes, such as when it enters a region with a different electric field or if it interacts with other particles.