A particle beam is made up of many protons each with a kinetic energy of 3.25times 10-15J. A proton has a mass of 1.673times 10-27kg and a charge of +1.602times 10-19C. What is the magnitude of a uniform electric field that will stop these protons in a distance of 2 m?

I am struggling with this problem. I don't understand where I use the KE or mass. I tried doing E=kq/r^2 but the answer I got was wrong.

Force = charge * electric field

or
F = Q E
if you do F = m a you can use the mass
but you can use
work done = change in Ke

work done = change in Ke= force * distance

so
Q E d = Ke

2 meters= 3.25*10^-15/(1.602^10^-19 E)

thanks!I also have a question about this one.

When a test charge q0 = 9 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction.

(c) If this force is due to a charge on the y axis at y = 3 cm, what is the value of that charge?

so in a I found E to be 88888.889
in b I found the F on a charge of -4nC to be -3.555e-4

I thought for C I would just use E=kq/(r^2) but it was wrong

To solve this problem, we need to understand the relationship between the kinetic energy of a proton, the distance it travels, and the electric field strength acting upon it.

First, let's recall the formula for kinetic energy (KE) of an object:

KE = (1/2) * m * v^2

Where:
KE is the kinetic energy,
m is the mass of the object, and
v is the velocity of the object.

Now, we also know the formula for the electric force (F) experienced by a charged particle in an electric field (E):

F = q * E

Where:
F is the force experienced by the particle,
q is the charge of the particle, and
E is the electric field strength.

In this case, we want to find the magnitude of the uniform electric field (E) that will stop the proton in a distance of 2 meters. To stop the proton, the electric force acting on it must be equal in magnitude but opposite in direction to the force created by its kinetic energy.

To solve the problem, we need to equate the force experienced by the proton due to the electric field to the force required to stop the proton. We can set up the equation as follows:

F (electric field force) = F (force required to stop the proton)

q * E = m * a

Where:
a is the acceleration of the proton.

The proton's acceleration can be calculated using Newton's second law of motion:

F = m * a

Rearranging the formula, we get:

a = F / m

Substituting the original equation, we have:

q * E = (m * a)

E = [m * a] / q

Now, we know that acceleration (a) is related to the distance (d) traveled by the object using the formula:

d = (1/2) * a * t^2

Since the particle stops after traveling a distance of 2 meters, we can rearrange the above formula for acceleration:

a = (2 * d) / t^2

Substituting this value for acceleration into our equation for electric field strength (E), we get:

E = [m * ((2 * d) / t^2)] / q

Finally, let's substitute the given values:

m = 1.673 × 10^-27 kg
d = 2 m
t = ?
q = 1.602 × 10^-19 C

We can solve for t using the kinetic energy (KE) formula:

KE = (1/2) * m * v^2

Rearranging the formula and substituting the given KE value:

v = sqrt((2 * KE) / m)

Substituting this value for velocity (v) into the formula for time (t), we have:

t = d / v

Now, let's substitute the given values:

m = 1.673 × 10^-27 kg
KE = 3.25 × 10^-15 J
d = 2 m

Substitute these values into the equation to solve for t. Then substitute the found value of t into the equation for electric field strength (E) to get the final answer.

This calculation will give us the magnitude of the uniform electric field needed to stop the protons in a distance of 2 meters.

Just use E = F/q