Electric Potential: Calculating the Final Ball Potential

[glueball8]

Homework Statement

Electrons are fired at a ball an infinite distance away at 100kev. The distance from the ball's center to the line in which the electrons are fired in is 0.6R. What is the ball's final potential at the radius?

Homework Equations

$$E=\frac{eQ}{4\pi \varepsilon_{o}\times R}$$

$$V=\frac{Q}{4\pi \varepsilon_{o}\times R}$$

The Attempt at a Solution

I'm think that 100keV*0.4 = keq/r therefore V=40kV. There's a lot of assumptions I'm not sure which ones I should make.

 

[anathema]

Hello,

Thank you for your post. It seems like you have a good start on solving this problem. However, there are a few things that need to be clarified and addressed.

First, the equations you provided are correct, but they are for calculating the electric field and potential at a point due to a point charge. In this problem, we are dealing with a charged ball, not a point charge. So we need to use a different equation that takes into account the size and shape of the ball.

Second, the units in your attempt at a solution are not consistent. The equation for electric potential has units of volts (V), not keV. So we need to convert 100 keV to volts before using it in our calculations.

Finally, you mentioned making assumptions, which is a good point to address. In physics, we often have to make simplifying assumptions in order to solve a problem. However, it's important to make sure that these assumptions are reasonable and do not significantly affect the accuracy of our solution.

Now, let's look at how we can solve this problem. We are given the distance from the center of the ball to the line in which the electrons are fired (0.6R) and the initial energy of the electrons (100 keV). We are also told that the ball is an infinite distance away, which means that we can assume the electric potential at infinity is zero.

To solve for the final potential at the radius, we need to use the equation for the electric potential due to a charged sphere:

V = kQ/R

Where k is the Coulomb constant (9x10^9 Nm^2/C^2), Q is the charge of the sphere, and R is the distance from the center of the sphere.

To use this equation, we need to find the charge of the sphere. We can do this by using the initial energy of the electrons and the equation for the electric potential:

V = kQ/R

Q = VR/k

Q = (100 keV)*(1.6x10^-19 C)/(9x10^9 Nm^2/C^2)

Q = 1.78x10^-12 C

Now, we can plug in this value for Q and the given value for R (0.6R) into the equation for electric potential to find the final potential at the radius:

V = (9x10^9 Nm^2/C^2)*(1.

 

[nlightner]

I would approach this problem by first clarifying some of the assumptions that have been made. For example, it is not specified whether the ball is positively or negatively charged, and whether the electrons are being fired in a straight line or in a curved trajectory.

Assuming that the ball is positively charged and the electrons are being fired in a straight line, we can use the equations provided to calculate the final potential at the radius. First, we can calculate the electric field at the radius using the equation E = (eQ)/(4πεoR), where e is the charge of an electron (1.6 x 10^-19 C), Q is the charge of the ball, and R is the distance from the ball's center to the line in which the electrons are fired (0.6R in this case).

Next, we can use the equation V = (Q)/(4πεoR) to calculate the potential at the radius, where V is the potential, Q is the charge of the ball, and R is the distance from the ball's center to the radius.

It is important to note that the 100keV given in the problem statement is the kinetic energy of the electrons, not the potential. Therefore, we cannot simply multiply it by 0.4 to get the potential. We need to take into account the charge of the electrons and the distance they are being fired from, as shown in the equations above.

In conclusion, without knowing the specific charge of the ball and the exact trajectory of the electrons, it is not possible to calculate the final potential at the radius. However, by clarifying these assumptions and using the equations provided, we can make a more accurate calculation.