<p>When a positive charge +q moves in a uniform electric field, then it does positive work.
When a negative charge -q moves in a uniform electric field, then there's negative work by the electric field pushing against it.</p>
<p>Negative charges are accelerated by electric fields toward points
(A) at lower electric potential
(B) at higher electric potential</p>
<p>So, if the equation for electric potential is: Change in V = Change in Ue / q and We is negative, then Change in V = -(We)/-q = A Negative number. So change in v is negative, which means electric potential went down. But the answer is B.</p>
<p>Which points in the uniform electric field shown above lie on the same equipotential?
Answer: 4 and 2</p>
<p>But is it fair to say that 1 and 3 have the same charge magnitude? Or is it wrong? If I saw two capacitors then I would know where the same voltage is (if they're parallel), but it seems like there's only one parallel-plate capacitor here, and I don't know what to do.</p>
<p>Keep in mind that in Physics, the reference is always using positive charges; e.g. current is the flow of positive charge and electric field lines are drawn to show how positive charges would flow.</p>
<p>With that said, positive charges ALWAYS flow from high to low potential, which, logically, should make a decent amount of sense. However, if a charge is negative it goes the opposite, from low to high potential - there’s the answer B.</p>
<p>On the next one, you need to know the equation for a parallel plate capacitor, (Epsilon * A)/d. You can tell that capacitance depends on distance, and using the equation C = Q/V (Q will be the same for both plates, keep that in mind) you know that if C increases, voltage must decrease. Not 100% sure about that #2 explanation but the first one is certain.</p>
<p>For number 2, I think d refers to the distance from the plate. So, in a horizontal line in between the plates, d doesn’t change, therefore C doesn’t change. Thus, V is the same across that line, and so, field line is equipotential.</p>
<p>I understand your first explanation, but how about the formula: V = U/q => - (-qEr)/-q = V would be negative, which gives (A) as the answer.
Could you help me out with this?</p>
<p>The question itself says that negative work is done when the negative particle is accelerated; U=-W thus if negative work is done there is an increase in potential.</p>