What Is The Strength Of The Electric Field At The Position Indicated By The Dot In (figure 1)?11/13/2019
Figure 1Figure 2INTRO:TACTICS BOX 32.1 Right-hand rule for fields. Point your right thumb in the direction of the current.(Figure 1). Curl your fingers around the wire to indicate a circle.
Your fingers point in the direction of the magnetic field lines around the wire.PART A: The following sketches show a wire carrying a current I in the direction indicated. Which sketch correctly shows the magnetic field lines around the wire?a) b) c)SOLUTION:First off, we know that the magnetic field can't be going two different directions, such as it is in b), which rules out that answer. Next we must use the right hand rule to determine the direction it goes. Pointing our right thumb downward, in the direction that I points, our fingers wrap counterclockwise around the bar. Since a) demonstrates this, we know that the correct answer is a).PART B: The magnetic field lines around a current-carrying wire are shown in the figure. In what direction is the current flowing in the wire?
What Is The Strength Of The Electric Field At The Position Indicated By The Dot In (figure 1)Problem 20.25 - Enhanced - With Feedback
(Figure 2)SOLUTION: For this one, pay special attention to which of the blue lines are in front of the diagram of the rod and which ones are behind, because this determines the direction of the magnetic field and also how to determine the direction of the current. So, that being said, place a right hand in the air and curl your fingers in the direction of the field, towards you as if you're looking at your fingernails. This results in the thumb pointing to the right. This means that this is the direction that the current is flowing in the wire. So the final answer for this part is ' from left to right '.
What Electric Field E⃗ (strength) Will Cause The Bead To Hang Suspended In The Air?
INTRO: The figure is a graph of E x. The potential at the origin is -250 VPART A:What is the potential at x = 3.0 m?Express your answer to two significant figures and include the appropriate units.SOLUTION:To determine the potential at the given point, we need to determine the potential difference and add the initial potential. INTRO: The electric potential along the x-axis is V=100e −2x/m V, where x is in meters.PART A: What is E x at x=1.0m?Express your answer as an integer and include the appropriate units.SOLUTION:E = -dV(x)/dx, the derivative of V is - 100e −2x/m ⋅(-2) V/m = 200e −2x/m V/mE(1) = 200e −2 V/m = 27.067 V/m ≈ 27 V/mPART B: What is E(x) at x=1.6m?Express your answer to two significant figures and include the appropriate units.SOLUTION:E = -dV(x)/dx, the derivative of V is - 100e −2x/m ⋅(-2) V/m = 200e −2x/m V/mE(1.6) = 200e −2⋅1.6 V/m = 200e −3.2 V/m = 8.2 V/m. − E ( y 1 − y 2NOTE:the expression E ( y 2 − y 1 ) will not yield the correct potential if you apply it to two points on opposite sides of the sheet.
For example, the expression does not indicate that two points on opposite sides of the sheet and the same distance from it are at the same potential ( V A B = V A − V B = 0 ), which is clear from the symmetry of the situation. If you take care in carrying out the integration to observe the change in the direction of the electric field as you pass from one side of the sheet to the other, you will find that the potential difference between A and B is actually given.
The problem statement, all variables and given/known dataa) What is the strength of the electric field at the position indicated by the dot in the figure?b) What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.Figure attached2.
Relevant equationsE = Kq/r^23. The attempt at a solutionAt first I thought strength of the electric field on both side would be same and the net strength would be Othen I calculated the distance from the point to the charge using Pythagoras theorem and then using the above equation which give me E as 1836 N/C.
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So for both side it would be 1836.2 = 3673 N/Cbut the answer was wrong?where did I mess up?
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