A scientist measures a 3.00-mV rms voltage across a 1.50-m-lon...

Question

A scientist measures a 3.00-mV rms voltage across a 1.50-m-long sensor that is aligned with the electric field of an electromagnetic wave. What is the electric field strength and the corresponding rate of energy transport per m²?

Accepted Answer

Everyone. Let's take a look at this practice problem done with electromagnetic waves. When this problem, a scientist measures a 3.0 millivolt R MS voltage across a 1.5 m long sensor aligned with an electromagnetic waves. Electric field. There are two parts of this problem. For part one, what is the electric field strength? And for part two, what is the corresponding rate of energy transport per square meter? Bringing four possible choices as our answers. Choice A for part one, the electric field is equal to 2.0 millivolts per meter. And for part two, the rate of energy is equal to 1.1 multiplied by 10 to the negative eight watts per meter squared. For choice B for part one, the electric field is equal to 3.0 millivolts per meter. And for part two, the rate of energy is equal to 2.3 multiplied by 10 to the negative eight watts per meter squared. For choice C for part one, the electric field is equal to 2.0 millivolts per meter. And for part two, the rate of energy is equal to 4.5 multiplied by 10 to the negative eight watts per meter squared. And for choice D for part one, the electric field is equal to 3.0 millivolts per meter. And for part two, the rate of energy is equal to 6.8 multiplied by 10 to the negative eight watts per meter squared. So for part one, we need to calculate the electric field strength. And for this, we need to recall our relationship between the electric field and the voltage. And so we call that relationship. We're gonna have er MS is equal to VR MS divided by D where er MS is the um R MS uh strength of the electric field. The VRS is the R MS voltage across the sensor. And D here is the distance that that voltage has been measured across which in this case will just be the length of the sensor. We have values for both the VR MS and the D we can plug those in. So we'll have er MS is equal to or VR MS that is three millivolts and this could be divided by the length of the sensor which is 1.5 m. And so when I plug those into my calculator, I'll get er MS is equal to 2.0 millivolts per meter. This can be our answer for part one and we left it in millivolts because our answer choices are in millivolts per meter. Now, for part two, we need to find the rate of energy transport per square meter. And that by definition is gonna be the magnitude of our pointing vector to recall your formula for the pointing vector in terms of the electric field that is going to be S is equal to C multiplied by epsilon, not multiplied by er MS squared or C. Here is gonna be the speed of light epsilon nut is the primitivistic of free space and S is our magnitude of our pointing vector. We actually have values for everything on the right hand side, since C and the epsilon knot are constants. So we'll have S is equal to the speed of light and vacuum. That is three multiplied by 10 to the eight meters per second. This could be multiplied by epsilon knot which is 8.85 multiplied by 10 to the negative 12, call them squared heard Newton meter squared. And then we just calculated our R MS um electric field. But we're gonna need that in volts per meter instead of millivolts per meter. So we'll need to multiply our answer by 10 to the negative three to convert millivolts into volts. We'll have two multiplied by 10 to the negative three that will give us, give us units of volts per meter. And that quantity is gonna be squared. And so when I plug those values into my calculator, I will get S is equal to 1.1 multiplied by 10 to the negative eight watts per meter squared here have just kept two significant figures. Let's give our answer for part two. So now if I look at my answer choices, this actually corresponds to answer choice. A. So the trick to this problem was just to realize that you needed to use your relationship for the electric field between the electric field and the voltage for part one. And for part two, you needed to use your definition for the magnitude of the pointing vector that the pointing vector is the rate of energy transport per square meter. And once you realized that those were two equations that you needed, it was just simply plugging in the values that you were given in the problem. So I hope that this has been useful and I'll see you in the next video.

A scientist measures a 3.00-mV rms voltage across a 1.50-m-long sensor that is aligned with the electric field of an electromagnetic wave. What is the electric field strength and the corresponding rate of energy transport per m²?

Key Concepts

RMS Voltage

Electric Field Strength

Energy Transport Rate

Watch next

A scientist measures a 3.00-mV rms voltage across a 1.50-m-long sensor that is aligned with the electric field of an electromagnetic wave. What is the electric field strength and the corresponding rate of energy transport per m2?

Key Concepts

RMS Voltage

Electric Field Strength

Energy Transport Rate

Watch next

A scientist measures a 3.00-mV rms voltage across a 1.50-m-long sensor that is aligned with the electric field of an electromagnetic wave. What is the electric field strength and the corresponding rate of energy transport per m²?