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In our lecture, we are currently learning about electric potential and electric potential energy.

However, I am very confused by the correct use of signs for electric potential.

In the class, we learned that the change in potential difference is equal to $V= -Ed\cos(\theta)$. However, when solving some textbook questions, they somehow seemed to have neglected the negative sign, and just calculated the potential energy/electric field at a point to be positive. i.e. they used the equation $V=Ed\cos(\theta)$ without the negative sign.

Anyone know why??

EDIT: Here is the question in my book I was talking about. The answer is apparently 1.5*10^6 V/m enter image description here

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  • $\begingroup$ Can you please please give us that problem where your book has used it differently? $\endgroup$
    – user240696
    Feb 21 '20 at 15:42
  • $\begingroup$ Yes sorry, I've just added it now. $\endgroup$
    – Luchardo
    Feb 21 '20 at 15:59
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When we have two charged plates (let's assume for the sake of explanation that one is negatively charged and the other is equally positively charged) the direction electric field is from positive plate to negative plate and the picture looks something like this

enter image description here

Now, work done on unit charge to move it from negative plate to positive plate is what we call potential difference and therefore, we can write $$ -\int_{negative}^{positive} \mathbf E \cdot d\mathbf s = \Delta V$$ the key point here is that potential is defined as the work done against the field and negative plate is at a lower potential and positive plate is at higher potential therefore potential difference in this setup will be positive, but notice that we are moving from negative plate to positive plate therefore distance travelled will be negative 1cm (as we are moving left and in Cartesian system it is the negative direction)$$-\int_{negative}^{positive}E ~ds = 1.50 \times 10^4 V$$ $$ -E \int_{negative}^{positive}ds = 1.50 \times 10^4 V $$ $$ -E \times (-10^{-2}) = 1.50 \times 10^4 V$$ Therefore, $$ E = 1.5 \times 10^6 $$ .

Hope this helps! If you have any doubt you're welcome to ask it in comments.

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  • $\begingroup$ WOW, thank you so much for the GREAT explanation!! So basically, the potential difference is the work needed to move a positive charge from the negative to the positive plate. Therefore, is it correct to assume that for any parallel plates, the potential difference is negative, and therefore the negative sign in the equation disappears? $\endgroup$
    – Luchardo
    Feb 21 '20 at 16:35
  • $\begingroup$ @Luchardo Well your last sentence is quite sloppy. Potential difference is difference of potentials at final point and the potential at the initial point. $\endgroup$
    – user240696
    Feb 21 '20 at 16:51
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    $\begingroup$ @Luchardo Thank you so much, your comment has helped me in figuring out the mistake in my answer. Actually, when evaluates the integral I put the $10^-2$ without the minus sign. You see we are following sign rules so if we are moving left we must do the subtraction accordingly. $\endgroup$
    – user240696
    Feb 21 '20 at 17:42

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