Calculating e/m Ratio: Step-by-Step Guide for Accurate Results

[barcelonaboy]

Homework Statement

I'm trying to calculate the charge ratio of e/m, I've plotted the given data points to get a straight line graph with a gradient of 0.848.

Homework Equations

I know that I can calulcate the charge ratio of e/m by using
e/m = (4.49*10^8 * gradient^2)/volts

where volts = 3000.

The Attempt at a Solution

When I plug the numbers in I get a value of 107625 which is nowhere near the given book value of e/m 1.7*10^11 Ckg^-1

I think that I'm doing something in the wrong units as I need to get the answer in C/kg^-1, but I can't see where I'm going wrong.

Any pointers would be much appreciated

 

[Jhero]

.

it is important to always check your units and ensure they are consistent throughout your calculations. In this case, it seems like you are using the correct equation but may have made a mistake in converting units.

First, let's look at the units for e/m. The unit for e/m is C/kg^-1, which means Coulombs per kilogram. This tells us that the unit for e/m should be in terms of charge per unit mass.

Next, let's look at the units for the given data points. The gradient is given in terms of meters per second squared (m/s^2). However, in order to use the equation, we need to convert this to units of volts. This can be done by multiplying the gradient by the distance between the plates (in meters) and dividing by the voltage (in volts). This will give us units of volts/meter, which is the correct unit for the gradient in this equation.

Now, let's look at the units for the voltage. The given voltage is 3000 volts, which is the correct unit for voltage. However, in order to use this in the equation, we need to convert it to units of meters. This can be done by multiplying the voltage by the distance between the plates (in meters) and dividing by the gradient (in m/s^2). This will give us units of meters, which is the correct unit for voltage in this equation.

Finally, let's put everything together. We have the gradient in units of volts/meter, the voltage in units of meters, and the given value for e/m in units of C/kg^-1. When we plug these into the equation, we get a value for e/m in units of C/kg^-1, which is consistent with the given book value.

In summary, it is important to always check your units and make sure they are consistent throughout your calculations. In this case, the mistake was most likely a unit conversion error. By double checking your units, you can ensure that your final answer is in the correct units and accurately reflects the given data.