ExamKrackers 1001 physics #773

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patrickmbyrne

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Tried searching for it. Here they explain that work will only occur moving parallel or antiparallel to the electric field lines. So the answer they give makes sense to me, but I'm not sure why work is being done in a constant electric field. Does constant not mean equal potential throughout the field? i.e. Work moving from a to b = 0.
 
Hi, I think you are mixing up "constant electric field" and equipotential lines. The electric field is constant here but perpendicular to the equipotential lines (where the electrical potential energy is the same) . Assume that it is a positive test charge and the electric field lines pointing right to left are going from higher potential energy to lower potential energy (the test charge has more potential energy when it is near the "positive" side on the right because it repels positive charge and naturally wants to move towards the negative side on the left).

In the example, you are doing work to move the charge from lower potential energy to higher potential energy by moving it to the right a certain distance against its "natural direction" which is to the left... The answer would be Edcos(angle).

To summarize, the test charge wants to move left but you are moving it to the right a certain distance instead. So you find how much it moved to the right. Since it is moving at an angle then you calculate that using cos
 
Hi everyone! I was also wanting some clarification on this one. I'm not sure if I oversimplified this question, but I simply looked at it based on my understanding of work from kinematics. Essentially, if there is displacement in the direction of the force, then work is done on an object. And we know that since electric field dictates electric force, in this case, there is a force present, since there is a constant electric field present. Did I make it too easy? Or is this all that we were supposed to recognize in this question? Or am I missing something?
 
@ohitsthatguy: the question is basically summarized in the image I've attached in this reply. Based on that image, the question asks "what is the change in voltage experienced by the charge q?"
 

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Yeah I was thinking that anywhere in a constant electric field would have equal potential. I don't know why this thing didn't alert me when people replied to the thread. Guess your answers make sense I just have to remember that constant electric field does not equal constant potential.
 
Yeah I was thinking that anywhere in a constant electric field would have equal potential. I don't know why this thing didn't alert me when people replied to the thread. Guess your answers make sense I just have to remember that constant electric field does not equal constant potential.

I hope I'm on the right track. I'm no genius. If anyone can clarify some of my doubts listed in my posted, that would be much appreciated as well!
 
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