The potential difference of a proton

[Johnnie123]

1. Homework Statement
A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton:

a. Remains at rest at B.
b. Moves toward A with a steady speed.
c. Moves toward A with an increasing speed.
d. Moves toward C with a steady speed.
e. Moves toward C with an increasing speed.

Homework Equations

$$U = qV$$

The Attempt at a Solution

The correct answer is c, but I am having a hard time understanding why. Charged particles may increase or decrease in speed if they are in an area with an electric potential. In this case, we have a potential difference - -100 V to 100 V - but how do I determine what $V$ is equal to. The potential difference could be positive in one sense and negative in the other.

 

[DrClaude]

In the absence of other information, you should assume that the potential varies linearly between the different points.

 

[ehild]

View attachment 238035


The correct answer is c, but I am having a hard time understanding why. Charged particles may increase or decrease in speed if they are in an area with an electric potential. In this case, we have a potential difference - -100 V to 100 V - but how do I determine what $V$ is equal to. The potential difference could be positive in one sense and negative in the other.

$V$ means "Volt" unit of voltage and electric potential.100 V is 100 volt potential, -100 V is -100 volt potential.

 

[kuruman]

In the absence of other information, you should assume that the potential varies linearly between the different points.

I think that's a fair assumption. It looks like the dashed lines are equipotentials.

 

[ZapperZ]

View attachment 238035

1. Homework Statement
A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton:

a. Remains at rest at B.
b. Moves toward A with a steady speed.
c. Moves toward A with an increasing speed.
d. Moves toward C with a steady speed.
e. Moves toward C with an increasing speed.

Homework Equations

$$U = qV$$

The Attempt at a Solution

The correct answer is c, but I am having a hard time understanding why. Charged particles may increase or decrease in speed if they are in an area with an electric potential. In this case, we have a potential difference - -100 V to 100 V - but how do I determine what $V$ is equal to. The potential difference could be positive in one sense and negative in the other.

Note that the more accurate relationship here is

$$\Delta U = q\Delta V$$

So it is the change in V, not the absolute value of V, that is important. Next, the electric force here is due to the presence of electric field E. This is where you need the relationship between E and V, i.e.

$$E = \frac{\Delta V}{\Delta x}$$

E is the gradient of V (taking into account just the magnitude). Again, it is the change in V over a distance that is important here.

So look at the diagram, and see whether you can identify the direction of where the E field should point, and then find if you can understand why there actually is a force acting on that charge.

BTW, this looks like a quick question out of Pearson's Mastering Physics.

Zz.