How Does Electric Field Direction Relate to Equipotential Lines?

AI Thread Summary
In the discussed scenario, equipotential lines are vertical and parallel to the y-axis, indicating that the electric field must point in the x-direction. Since the potential increases in the positive x-direction, the electric field vector will point in the negative x-direction due to the relationship E = -dV/dx. The potential is higher near the positive charge, but the electric field always points towards lower potential, which is near the negative charge. The confusion arises from understanding the signs in the equations relating electric field and potential. Ultimately, the electric field direction is confirmed to be negative in the x-direction.
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Homework Statement


The equipotential curves in a certain region of an equipotential diagram in an xy plane are parallel to the y-axis and are equally spaced. The potential is increasing in the +x direction. Which direction would an electric field vector point in this region?

Homework Equations


N/A


The Attempt at a Solution


The equipotential lines are vertical, so the electric field lines must be in the x-direction since they run perpendicular to each other. Would they point in the positive x direction since the potential is increasing there? I'm having a hard time picturing what this would look like.
 
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Try to relate electric field and potential/potential difference- is there any formula by which potential can be written in terms of electric field? Think about it and let me know if you have any question.

Best of luck!
 
I know that V=∫Edl, but I'm having trouble connecting that to what the equipotential lines would look like.
 
bbuilder said:
Would they point in the positive x direction since the potential is increasing there?
Suppose the field is created by some distant pair of opposite charges, one in positive x direction and one in the negative. Will the potential be higher towards the positive charge or towards the negative? Which way will the field point?
I know that V=∫Edl
No, it's V=-∫Edl. In differential form: E = -dV/dl (or -dV/dx in this context). You're told the potential increases with increasing x, so what is the sign of dV/dx?
 
haruspex said:
Suppose the field is created by some distant pair of opposite charges, one in positive x direction and one in the negative. Will the potential be higher towards the positive charge or towards the negative? Which way will the field point?

No, it's V=-∫Edl. In differential form: E = -dV/dl (or -dV/dx in this context). You're told the potential increases with increasing x, so what is the sign of dV/dx?

The potential would be the same at both a positive charge and a negative charge because potential is scalar. The electric field would point towards the negative charge.

The sign of dV/dx is positive, but when the negative sign is taken into account then E is negative, so the electric field is going in the negative x-direction. Is this correct?
 
bbuilder said:
The potential would be the same at both a positive charge and a negative charge because potential is scalar.
Eh?! Is +3V the same as -3V?
The electric field would point towards the negative charge.

The sign of dV/dx is positive, but when the negative sign is taken into account then E is negative, so the electric field is going in the negative x-direction. Is this correct?
Yes.
 
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