A 10-cm-long thin glass rod uniformly charged to +10 nC and a ...

Question

A 10-cm-long thin glass rod uniformly charged to +10 nC and a 10-cm-long thin plastic rod uniformly charged to −10 nC are placed side by side, 4.0 cm apart. What are the electric field strengths E₁ to E₃ at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?

Accepted Answer

Hello, let's go through with this practice problem. Two thin bars of length, 12 centimeters are placed vertically in the same plane with a distance of six centimeters between their centers. The bars centers lie on the same horizontal line. The left bar is uniformly charged to 0.1 five micro cools. And the right is uniformly charged to negative 0.15 micro cools determine the magnitude of the electric field at a point located on the line connecting the midpoints of both bars, which is positioned two centimeters to the right of the left bar option. A 1.5 multiplied by 10 squared noons per Coolum B 3.2 multiplied by 10 to the power of three new and per Coolum C or multiplied by 10 to the power of four and D 1.5 multiplied by 10 to the power of six. OK. So this problem might seem a little confusing at first. So let's first begin by drawing a diagram of the situation. We have two bars, one on the left side and one on the right side. OK. And they are separated by a distance of six centimeters and the length of the bar, the length of both bars is 12 centimeters. We're told that both bars have the same charge magnitude, but the left one is positively charged while the right one is negatively charged. Also recall that the convention for electric field is that electric field lines point away from positive charges and towards negative charges. So because the left hand side of this system shows positive charge, then there are going to be electric field lines pointing away from the left bar and the right bar is negatively charged. So there are going to be field lines pointing towards it. So between these two bars in the specifically in the region that we're looking for the electric field is pointing from left to right. What this problem is actually asking us for it is asking for us to consider a line that passes through the midpoint of both bars. Kind of like this. The problem's asking us to find the magnitude of the electric field at a point on this line that is two centimeters to the right of the left bar. So this points right here. Now recall that we have been provided with a formula or the magnitude of the electric field strength within the plane that bisects a charged rod, that formula states that the electric field is equal to one divided by four pi epsilon knot multiplied by the magnitude of the charge on the bar divided by R where R represents the distance between the point and the charged bar generating the field multiplied by the square root of our squared plus the square of L divided by two, where L represents the full length of the bar. So all we need to do in this problem is plug in the values that were given to us in the equation. But there is an important thing for us to remember there is an electric field being generated by two different bars. So there are two different components of the electric field to consider. If we're looking to find the net electric field at that point, we need to consider the effects created by both bars. So what we'll actually do is say that the electric field at that point is the sum of the electric fields from each bars. So let's start with the left bar. So the left for the left bar and that's one divided by four pi epsilon knot multiplied by the magnitude of the left charge. And it's the same for both both bars. But the magnitude of the left charge divided by the distance away from the left bar, which is give, which is given in the problem as two centimeters multiplied by the square root of R sub L squared plus the square of half the length. And then to this, we're going to add the same formula but with the variables that apply to the rightmost bar. So plus one divided by four pi epsilon knot multiplied by the magnitude of the rightmost charge multiplied by the distance between that point and the rightmost bar multiplied by the square root of R sub R squared plus the square of half the length. Since we have two different R values, now let's quickly establish what those are because for our sub L of course, that's just the two centimeter difference between the point and the leftmost bar. But since the total distance between the two bars is six centimeters, that means that our sub R, the distance between the point and the rightmost bar is going to be six centimeters minus two or four centimeters. So now we're ready to plug everything into our equations. So I'm gonna factor out the one divided by four pi epsilon knot for simplicity and the electric field is equal to the leftmost charge magnitude which is 0.15 multiplied by 10 to the power of negative six coons divided by the distance between R. It's R sub L. So that's two centimeters or 0.02 m multiplied by the square root of R sub L squared. So 0.02 m squared plus the square of half of L or which is 12 centimeters or 0.12 m divide by two and square it. And then we'll add the term for the rightmost bar. So the magnitude of the rightmost bar is again, 0.150 multiplied by 10 to the power of negative six columns, technically, it's negative, but we're only concerned with magnitudes and then R sub R is four centimeters as we just discussed. So that's 0.04 m multiply by the square root of R sub R squared. So that's 0.04 m plus the square of half the length. So 0.12 m divided by two squared. And if we put all of this into a calculator, then we find a net electric field magnitude of 1.5 multiplied by 10 to the power of six newtons for Coolum. And so that is our answer which corresponds to option D. So option D is the answer to this problem. And that is it for this problem. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

Key Concepts

Electric Field

Superposition Principle

Coulomb's Law

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