Question Video: Finding the Charge Contained within a Volume given the Electric Field

The electric field in a region is given by 𝐄 = (π‘Ž/(𝑏 + 𝑐π‘₯))𝐒, where π‘Ž = 200 Nβ‹…m/C, 𝑏 = 2.0 m, and 𝑐 = 2.0. What is the net charge enclosed by the shaded volume shown?

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Video Transcript

The electric field in a region is given by 𝐄 equals π‘Ž divided by 𝑏 plus 𝑐π‘₯ in the 𝐒-direction, where π‘Ž equals 200 newton-meters per coulomb, 𝑏 equals 2.0 meters, and 𝑐 equals 2.0. What is the net charge enclosed by the shaded volume shown?

We can call this net enclosed charge capital 𝑄. And to start solving for it, we can recall a mathematical relationship between charge and electric field 𝐄. Gauss’s law tells us that if you have a closed surface, a shape with some volume and area, then if you add up that entire area multiplying each part by the electric field that goes through it, then that surface integral over a closed surface is equal to the charge enclosed by the surface over the constant πœ– naught, the permittivity of free space. We’ll treat that constant as having an exact value of 8.85 times 10 to the negative 12 farads per meter.

As we consider how to apply this relationship to our scenario, we take a look at our shape, our surface, which is a box. The box has a height of one and a half meters, a width of 1.0 meters, and a length of 2.0 meters. Also the box is a closed surface. So that means it’s a candidate to use Gauss’s law to solve for the enclosed charge 𝑄. Let’s take a closer look at the dot product of the two vectors involved in this calculation. We have a vector, the electric field, dotted with an area element, 𝑑𝐴, which is also a vector which will point a normal or perpendicular to the area element it represents.

We know that the electric field in this example only points in one direction, in the positive 𝐒-direction, which means we can label our axes π‘₯, 𝑦, and 𝑧 for the vertical. What about the area elements of our box? Which ways do their area vectors point? Like any box, ours has six faces, each with its own area vector pointing normal to that face. Now we can recall that the dot product between two coplanar vectors, we’ll call them 𝐴 and 𝐡, is equal to the product of their magnitude times the cosine of the angle between them.

When we consider this dot product for our electric field, 𝐄, pointed in a positive 𝐒-direction and the six faces of our surface, since the cosine of 90 degrees is zero, that dot product for the four faces that don’t face in either the positive or negative π‘₯-direction is zero. So the only faces we need to account for are the front face, we’ll call that 𝐴 sub one, and then the back face, which we’ll call 𝐴 sub two. So our surface integral simplifies to two terms, the electric field 𝐄 first dotted with 𝐴 one plus the electric field 𝐄 dotted with 𝐴 two.

Now because the electric field is position dependent β€” that is, it has an π‘₯ variable in its equation β€” it can be helpful to write these two terms separately and calculate them separately to add them together to solve for 𝑄 over πœ– naught. We can calculate these two terms one by one. Let’s start with 𝐄 dot 𝐴 one. The magnitude of the area 𝐴 one is equal to 1.0 meter times 1.5 meters. And 𝐴 one points in the positive π‘₯- or positive 𝐒-direction. The electric field 𝐄 at this position is equal to 200 newton-meters per coulomb over 2.0 meters plus 2.0 times the π‘₯-coordinate of the plane 𝐴 sub one, which is 2.0 meters.

𝐄 also points in the positive 𝐒-hat or π‘₯- direction. So when we take the dot product of these two terms, we can simply multiply them together then find it’s equal to 50 newton-meters squared per coulomb. Next we move on to calculate 𝐄 dot 𝐴 two. In this case, the magnitude of the area is the same as before, but now it points in the negative 𝐒-direction. And 𝐄 now has a value of zero meters in its π‘₯- component because we’re in the 𝑧𝑦-plane. When we dot these values together, the result we find is negative 150 newtons-meter squared per coulomb.

When we add these two values on the left side of our equation together, we find that 𝑄 over πœ– naught is equal to negative 100 newtons-meter squared per coulomb. Or 𝑄 is equal to πœ– naught times negative 100 newtons- meter squared per coulomb. When we plug-in for πœ– naught and enter these values on our calculator, we find that 𝑄, to two significant figures, is negative 8.9 times 10 to the negative 10th coulombs. That’s the amount of charge enclosed in this volume.

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