Go to https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.ht

Question

Go to https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html. You can download the latest version of Flash here if needed, http://get.adobe.com/flashplayer/.

• Check “Show hi-res V” in the green box at the bottom right.
• Drag a “1 nC” red positive charge from the top right to the middle.
• Drag the rounded square box with the black crosshairs near the red positive charge. This box is a “V-field detector”. In the white box it shows the value of the V-field in the middle of the crosshairs.
* Question 1: Drag the detector in a circle around the positive charge keeping the crosshairs a constant distance from it. Describe the value of the V-field as you do this.
* Question 2: Drag the detector away further away from the positive charge. Describe the value of the V-field as you do this. Describe what the red color in the simulation is indicating.
Drag the positive charge back to the box at the top right. Drag a “1 nC” blue negative charge to the middle.
* Questions 3 and 4: Repeat questions 1 and 2 but with the negative charge present instead.
• Drag a positive charge to a location a moderate distance from the negative charge.
* Question 5: Use the detector to investigate the V-field produced by the two charges together. Describe your findings.
*Question 6: Investigate the region exactly halfway between the two charges. What is the value of the V-field in this region? Come up with a reason why this value makes sense.
*Question 7: Using any configuration of charges (your choice), investigate what the “plot” button on the detector does. Describe your findings.

Explanation / Answer

1)
Voltage indicates a constant value.
Voltage is given as V = kQ/r, r is the distance to the charge Q and k is a constant. When we keep the distance r the same, V will be a constant.

2)
When we drag the detector further away, the voltage decreases. This is in accordance with the above equation V = kQ/r, when r increases, V decreases. The red color indicates the intensity of electric field from positive charge and voltage.

3)
Since Q is negative, V = -kQ/r
detector will show a constant negative value

4)
As we drag the detector away, r increases and magnitude of V decrease.

5)
When we drag the detector away from the two charges, V falls off rapidly than that for a single charge. This is because for the combination of two charges (dipoles), V depends on the inverse square of the distance.

6)
V = 0 because the halfway between two equal and opposite charges is called equipotential surface.

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