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momowoo

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- Thread starter momowoo
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In summary, the direction of the field along the x-axis in a graph where V is a function of x depends on the slope of the potential graph. A positive slope corresponds to a negative field, pointing towards the -x axis (to the left), while a negative slope corresponds to a positive field, pointing towards the +x axis (to the right). This relationship is due to the field being related to the negative of the slope of the potential graph.

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momowoo

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Physics news on Phys.org

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berkeman

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Welcome to the PF.momowoo said:

Can you think of a charge distribution that might cause that V(x) graph? You might also consider if conductors might be involved...

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momowoo

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Would that be a hollow sphere?

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berkeman

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That's not what I'm picturing, but perhaps it might work as well. But it would take multiple hollow concentric conducting spheres charged to different voltages, no?momowoo said:Would that be a hollow sphere?

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Doc Al

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momowoo

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towards lower potential

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berkeman

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- #8

Doc Al

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Exactly. So what does that tell you about your diagram?momowoo said:towards lower potential

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momowoo

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- #10

Doc Al

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Where the graph shows a positive slope, the positive test charge would tend to move towards lower potential (to the left). So what direction must the field be in that region?momowoo said:

So how does the field direction depend on the slope of the potential graph?

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momowoo

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How do you define field direction? Wouldn't saying left and right be arbitrary?

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Doc Al

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Just specify the direction with respect to the x-axis given. ("To the right" corresponds to the +x direction, in this case.)momowoo said:How do you define field direction? Wouldn't saying left and right be arbitrary?

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momowoo

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momowoo

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Doc Al

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What's the direction of the field?

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momowoo

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To the left?

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momowoo

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- #18

Doc Al

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Yes. Where the potential rises to the right, the field points left.momowoo said:To the left?

- #19

momowoo

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THANK YOU SO MUCH FOR HELPING ME <3

- #20

Doc Al

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Yes. The field is related to themomowoo said:

You are very welcome.momowoo said:THANK YOU SO MUCH FOR HELPING ME <3

Electric potential refers to the amount of electric potential energy per unit charge at a given point in an electric field. It is a measure of the electric potential energy a charged object would possess at a specific point in the electric field.

While electric field measures the force exerted on a charged object at a specific point in the electric field, electric potential measures the potential energy that the charged object would possess at that same point. Electric potential is therefore a scalar quantity, while electric field is a vector quantity.

The unit of electric potential is volts (V), which is equivalent to joules per coulomb (J/C).

Electric potential is calculated using the formula V = kQ/r, where V is the electric potential in volts, k is Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the charge of the object in coulombs, and r is the distance from the object in meters.

Electric potential energy is the energy a charged object possesses due to its position in an electric field. It is directly proportional to the electric potential, with the equation U = qV, where U is the electric potential energy in joules, q is the charge of the object in coulombs, and V is the electric potential in volts.

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