# Finding the New Acceleration Given Point Charges

• Alex Z
In summary, a small sphere with a mass of 4.00×10−3 kg and a charge of +2.00μC is fired towards a fixed point charge with a speed of 40.0 m/s from a horizontal distance of 8.00 cm. The force of the system is calculated using F = (k*q*q) / (r^2), and using conservation of energy, the final distance of the sphere is found to be 0.0487 meters. The final kinetic energy is calculated to be 1.058, and using F = m*a, the acceleration is found to be 28144.5 m/s^2.
Alex Z

## Homework Statement

A point charge q1=15.00μC is held fixed in space. From a horizontal distance of 8.00 cm , a small sphere with mass 4.00×10−3kg and charge q2=+2.00μC is fired toward the fixed charge with an initial speed of 40.0 m/s . Gravity can be neglected.
What is the acceleration of the sphere at the instant when its speed is 23.0 m/s?

2. The attempt at a solution
I tried to use F = (k*q*q) / (r^2) to find the force of the system, which comes out to be 41.72 N. But when I try to use a = F/m, the answer is apparently wrong.

How did you calculate the force? (Where is the sphere when its speed is 23.0 m/s?)

Doc Al said:
How did you calculate the force? (Where is the sphere when its speed is 23.0 m/s?)
I calculated the force using everything was given to me. Since the radius of the sphere wasn't given, I just assumed that the size of the sphere was negligible and I therefore didn't have to take it into account.
As for the location of the sphere, I assume conservation of energy needs to be used to find the new distance?

Alex Z said:
I calculated the force using everything was given to me. Since the radius of the sphere wasn't given, I just assumed that the size of the sphere was negligible and I therefore didn't have to take it into account.
Sounds good.

Alex Z said:
As for the location of the sphere, I assume conservation of energy needs to be used?
Sure. If you show your calculations, we can look for possible errors. (Or confirm your results.)

Well I believe that the equation here would be Uei + Ki = Uef + Kf
So (k*q*q)/0.08 + 3.44 = (k*q*q)/rf + 1.137
With all this equated, rf should be equal to 0.0487 meters.

From there, my first instinct would be to find the change in distance and then use kinematics.

Show how you calculated the kinetic energies. (Check your arithmetic.)

Alex Z said:
From there, my first instinct would be to find the change in distance and then use kinematics.
Once you have the final position, you won't need kinematics.

Initial Kinetic Energy = 1/2*(4*10^-3)*(40^2) = 3.2
Final Kinetic Energy = 1/2*(4*10^-3)*(23^2) = 1.058

Oops. So that means that rf = 0.0487 meters. I guess the values weren't great enough to make a significant difference.

Now show how you calculated the force.

Doc Al said:
Now show how you calculated the force.

F = (k*q*q)/(r^2) = ((8.9*10^9)*(15*10^-6)*(2*10^-6)) / (0.0487^2) = 112.58 N

So from there, I can use F = m*a to find the acceleration:

a = F / m = (112.58) / (4*10^-3) = 28144.5 m/s^2

I believe that should be the final answer.

Much better!

Doc Al said:
Much better!

Awesome, thank you very much.

## 1. What is the formula for finding the new acceleration given point charges?

The formula for finding the new acceleration given point charges is F = qE/m, where F is the force, q is the charge, E is the electric field, and m is the mass of the charged particle.

## 2. How do you determine the direction of the acceleration?

The direction of the acceleration can be determined by the direction of the electric field. If the charge is positive, the acceleration will be in the same direction as the electric field. If the charge is negative, the acceleration will be in the opposite direction of the electric field.

## 3. Can the acceleration change over time?

Yes, the acceleration can change over time. This can happen if the electric field changes or if the charged particle moves closer or further away from the point charge.

## 4. What are the units for acceleration in this formula?

The units for acceleration in this formula are meters per second squared (m/s²).

## 5. How does the mass of the charged particle affect the acceleration?

The mass of the charged particle directly affects the acceleration. The larger the mass, the smaller the acceleration will be for a given force and electric field. Similarly, a smaller mass will result in a larger acceleration.

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