A −12 nC charge is located at (x, y)=(1.0 cm, 0 cm). What are the...

Question

A −12 nC charge is located at (x, y)=(1.0 cm, 0 cm). What are the electric fields at the positions (x, y)=(5.0 cm, 0 cm), (−5.0 cm, 0 cm), and (0 cm, 5.0 cm)? Write each electric field vector in component form.

Accepted Answer

Hello, everyone. Let's take a look at this practice problem. In a laboratory experiment, a tiny plastic sphere with a charge of negative eight nano columns is located at an X and Y position of two centimeters and zero centimeters. What is the electric field at the positions of point A with an X and Y position of zero centimeters and four centimeters determine the component form of the electric field vector at this position. We're given four multiple choice options. Option A 1.2 multiplied by 10 to the power of four na uh New Inra Coolum with an I hat vector plus 2.6 multiplied by 10 to the power of four nuance per Coolum along the J at Unter option B 1. multiplied by 10 to the power of four I nuance per Coolum minus 3.2 multiplied by 10 to the power of four J nuance per Coolum. Option C 4.2 multiplied by 10 to the power of four I nuance per Coolum plus 1.6 multiplied by 10 to the power of four J nuance per Coolum or option D 3.2 multiplied by the 10 to the power of four I nuance per colum plus 3. multiplied by 10 to the power of four J nuance per Coolum. OK. So this problem is actually not that bad. Basically, what's happening here is it's pretty simple. We have a charged particle shown here on the graph that has a charge, which means it is setting up an electric field all around it. The problem is simply asking us to find the in component form the electric field at a point some distance away from it. Note that point A is not a charged particle, it's just some point in space. And we're trying to find the electric field at that point. And nicely for us, we have a formula for the electric field at a distance away from a particle. It states that if you have a particle setting up an electric field, then that field is has a magnitude of the Coolum constant K multiplied by the charge on that particle. Using the magnitude form of the equation all divided by the square of R which is the distance from the charged particle to whatever point in space we're looking at for our purposes this arc, this R represents the distance between wherever the charge is and point A. So the first thing we want to do is find what that R distance is. You want to find the distance between the charge particle and position A. If we treat this setup as being kind of similar to a right triangle. Then we can use the Pythagorean theorem to find this value by taking the square root of the sum of the squares of the distances between them. Just from looking at the graph, we can see pretty easily that the horizontal distance between point A and charge Q is two centimeters or 0.02 m and the vertical distance between them is four centimeters or 0. m. If we put this Pythagorean theorem formula into a calculator, we can see that R has a value of about 0. m. So with that understood, let's plug that into our electric field formula to find the magnitude of the field at point A. So it's the Coolum constant 8.99 multiplied by 10 to the power of nine new and meter squared per colum squared multiplied by the magnitude of the charge or eight nano columns or eight multiplied by 10 to the power of negative nine columns, all divided by the square of the distance between the charge and point A or 0.04472 m. And this is squared. So if we put this into a calculator, then we find an electric field magnitude of 3.6 multiplied by 10 to the power of four new in per colum. Now this is the magnitude of the electric field at point A. But because the problem specifically asks us to break it up into component form. We're going to need to use some trigonometry to to break these components up. So the first thing I'm going to do is figure out how this is going to work. So this is point a right here and we want this broken up into components. So I'm going to define this angle here to be theta and it's an angle. We're going to need to know if we want to break this up into components. This is because from looking at the graph, we can see that the horizontal component, the X component of the field is going to be equal to the total magnitude of the field multiplied by the sign of that angle theta because the horizontal component is the opposite side from that angle of the triangle. Similarly, the Y component, the vertical component is equal to the magnitude of the field multiplied by the cosine of the angle theta. So the first thing we're gonna need to do is find that angle theta, we can use the tangent expression for this the tangent function. And so of course data is going to be equal to the inverse tangent of the opposite side, the opposite length of that right triangle or two centimeters divided by the adjacent component, the vertical component or four centimeters. And if we put this into a calculator and make sure your calculator is in degrees mode, then we find an angle of about 26.6 degrees. So that is our angle. So if you want to break this up into components, then we'll take our total magnitude 3.6 multiplied by 10 to the power of four and nuance per Coolum. And then multiply it by either the sine or the cosine of 26.6 degrees. So for the X component, we multiply the sign of 26.6 degrees to find a horizontal magnitude of 1. multiplied by 10 to the power of four new in per Coolum. Doing the same thing for the Y component, 3.6 multiplied by 10 to the power of four nuance per Coolum multiplied by the cosine of 26.6 degrees, put this into a calculator and this one is equal to 3. multiplied by 10 to the power of four new ones per Coolum. We have now found the magnitudes of the vertical and horizontal components. But we still need to work out the signs note that the charged particle is negative, which means that the electric field is going to be pointing towards the negative charge because electric field lines conventionally terminate at negative charges. And we can see from this arrow that, that means that the electric field is going to be positive in the X direction because it's pointing to the right but negative in the Y direction because it's pointing downwards. So if you were to write out a full expression for the electric field, then the horizontal component will be positive. So 1.6 multiplied by 10 to the power of four nuance per Coolum in the I hat direction. But then we'd have to subtract the Y component with the J hat unit vector. So minus 3.2 multiplied by 10 of the power of four new ones per Coolum with the J hat unit vector. And that is really it, that is a full component expression for the electric field. We found both components. And if you look at our multiple choice options, we can see that this agrees with option B. So option B then is the correct answer to the problem. And that is it for this video. I hope this video helped you out and I recommend checking out some of our other videos which will give you more experience with these types of problems, but that's all for now. And I hope you all have a lovely day. Bye bye.

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A −12 nC charge is located at (x, y)=(1.0 cm, 0 cm). What are the electric fields at the positions (x, y)=(5.0 cm, 0 cm), (−5.0 cm, 0 cm), and (0 cm, 5.0 cm)? Write each electric field vector in component form.

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