What is the electric potential energy U of the test charge at Point A? U = qV 6.

Question

What is the electric potential energy U of the test charge at Point A? U = qV 6. What is the electric potential energy of the test charge at Point B? 7. Has the electric potential energy of the lest charge increased or decreased in going from Point A to Point B? 8. What has happened to the kinetic energy KE of the particle in going from rest at Point A to whizzing pas Point B? 9. Let's assume we put a finger on the test charge and stop and hold it at Point B. If you now let go of the test charge, is the force it experiences (greater than, less than, or the same) as it was at Point A? Why? 10. What can you say about the strength of the electric field (E = F/q) at Point B compared to Point A?/(greater than, less than, or the same) 11. Sketch a few electric field lines in Fig. 4.7. 12. In the map of equipotential lines between two parallel plates, we see a simple relationship between the distance between plates and the change in potential. the entire distance from the positive plate to the negative plate is about 8 cm. the potential changes uniformly from 12 V to 0 V over that distance. This means that there is a change in potential of 12 V for every 8 cm or 1.5 V/cm. the test charge at A or at B or at any other point in this uniform field is said to be experiencing an electric field strength oi l.!> V cm. Write an equation relating electric field strength to potential difference and distance, tor the case oi tnc electric field produced by two parallel charged plates. 13. Would this same equation apply to the other fields you mapped? Why or why not? 14. We have been using the units V/cmfor electric field strength. Dividing by 100 would give the SI unit m Show (hat the V/mis equivalent to another SI unit for electric field strength, the N/C.

Explanation / Answer

12.) The electric field E and the potential difference V over a distance l are related by

|E| = - V/l = |V| / l

Note the usage of mod || here. Because the electric field is along the direction where the potential decreases. That is if the potential is decreasing from left to right (along the direction of E), the V (Vf - Vi ) will be negative. 0 - 12 = -12 in our case.

So, field E = - (-12)/ 0.08 = 150 Volts / meter

or simply forget the sign and use mod |E| = | V| /l = | -12| / 0.08 = 150 V /m

13.) Although I am not able to see the OTHER FIELD LINES you mapped, I can assure you that this equation applies for field lines between equipotential surfaces only (as the two plates of 12V and 0 V in our case) since we assumed that the potential changes uniformly with the distance while writing |E| = - V/l = |V| / l

14.) Potential is given by (1/4o ) x (q/r) and the SI unit for potential is Volt

Electric field is given by (1/4o ) x (q/ r2 ) observe the change in the denominator only. So, the SI unit for this Electric field is Volt / meter or V/m

Electric field when multiplied by charge gives electrostatic force F = Eq

And as we know the SI unit of force is Newton which is written as N simply.

So, writing F = Eq in terms of the units,   

units of force = (units of Electric field) x units of charge

N = (V/m ) x C (As we know the unit of charge is coulomb which is written as C)

=> N/C = V/m

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