Electrostatics

[dartmed]

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Questions:

If I have a positive charge and I am moving the positive charge against the electric field, how much work needs to be done to move the +q a distance d? Would it just be PE = qV assuming that I take the adjacent side of V if the charge is moved at an angle? If there is no V given, could I do V = E/L?

If a negative charge is moving against the electric field (like it's supposed to), is there any work done by the negative charge?

Thank you!!

 

[V5RED]

qV is for moving the charge from one end of a field to the other.

If you have E and q and d then you can use the common work equation of W=F*d

Recall that F=qE, so W=qEd.

 

[dartmed]

qV is for moving the charge from one end of a field to the other.

If you have E and q and d then you can use the common work equation of W=F*d

Recall that F=qE, so W=qEd.

What if the charge was moved at an angle? Would you take d cos theta? or would W just be qEd

 

[V5RED]

What if the charge was moved at an angle? Would you take d cos theta? or would W just be qEd

Assuming we are talking about a uniform electric field such as that between two oppositely charged plates, then the distance I am talking about refers to how much closer the charge gets to the positively charged plate. (ie how much it moves parallel to the field).

The angle is irrelevant since moving perpendicular to the field keeps you on an equipotential plane. The only use for the angle would be if they gave you the distance it moves and the angle between that motion and the field. You would then use that information to find the component of its displacement that is parallel to the field and treat that distance as d.

Also, for the negative charge example, the field does work on the charge.

 

[dartmed]

Assuming we are talking about a uniform electric field such as that between two oppositely charged plates, then the distance I am talking about refers to how much closer the charge gets to the positively charged plate. (ie how much it moves parallel to the field).

The angle is irrelevant since moving perpendicular to the field keeps you on an equipotential plane. The only use for the angle would be if they gave you the distance it moves and the angle between that motion and the field. You would then use that information to find the component of its displacement that is parallel to the field and treat that distance as d.

Also, for the negative charge example, the field does work on the charge.

That makes sense! Thank you!