When a positive charge is released and moves along an electric field line, it mo
ID: 1588549 • Letter: W
Question
When a positive charge is released and moves along an electric field line, it moves to a position of lower potential and lower potential energy lower potential and higher potential energy higher potential and lower potential energy higher potential and higher potential energy greater magnitude of the electric field. The electric potential inside a charged solid spherical conductor in equilibrium is always zero is constant and equal to its value at the surface decreases from its value at the surface to a value of zero at the center increases from its value at the surface to a value at the center that is a multiple of the potential at the surface is equal to the charge passing through the surface per unit time divided by the resistanceExplanation / Answer
Its definelty moving to a lower potential energy. PE = kQ*q/r
Where i chose r to be from the positive source of the elctric field outward. Notice if r increase your potential energy lowers. The positive chage has to move away from other potistive charges. The electric field will point from the postive source to the negative.
The potential is another for the voltage
The voltage is the potential energy per unit charge. Notice with the equation i just mentioned you can you can easily determine the sign.
PE = kQq/r where Q,q,k and r are all positive
If i divide by a negative charge to get the potential the potential will be negative and likewise for a postive charge
Therefore the potential at the postive side is postive and the negative side is negative (side meaing the electric field i chose is from two charged plates...a positive and negative side). So one side has a postive potenital while the other has a negative potential. Therefore the positive charge goes from higer to lower
lower potential and lower potential energy
2) the correct answer is b) is constant and equal to its value at the surface
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