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In a parallel plate capacitor, charge $q$ is travelling from negative to positive plate. Why is $\Delta V$ negative? Every book explains it by saying $\Delta U = -W$, but this equation came when we assumed that particle was travelling from higher potential to lower potential.

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Many forces $\vec{F}$ can be written as the gradient of some potential $V$. For example the force exerted to a particle of charge $q$ due to a potential $V$ is: $$\vec{F}=-q\,\vec{\mathrm{grad}}\,V \tag{1}$$ Note that $\vec{E}=-\vec{\mathrm{grad}}\,V$.

Because of the definition of this force, the potential $V$ is defined up to a constant. Set this constant without any loss of generality to $0$.

It is clear that the system in which this force is exerted, evolves until equilibrium of forces, $\textit{i.e.}$ $\vec{F}=0$ and there is no motion.

This stable equilibrium point can be reached only when the function $V$ has a minimum as you can see from $(1)$. Therefore the system always goes from high to low potential when no additional forces are exerted.

The work done by this particle through a curve $C$ is then: $$W=\int_{C}{\vec{F}\cdot d\vec{r}}=-q\int_{C}{\vec{\mathrm{grad}}\,V\cdot d\vec{r}} =-q\Delta V$$

Where $\Delta V$ is the total difference between the values of the potential evaluated at the extrema of the curve $C$.

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The minus sign in front of the integral will ensure that $\Delta V$ is positive in this case. The $\vec{E}$ between the plates of the capacitor points from the positively charged plate toward the negatively charged plate. If you're moving a charge from the negative plate to the positive plate, the $d\vec{r}$ points in the opposite direction of $\vec{E}$. So the dot product, $\vec{E} \cdot d\vec{r}$ is negative ($\vec{E}\cdot d\vec{r} = Edr\cos\theta$ but $\theta = 180^\circ$ so $\cos\theta = -1$). Then the minus sign in front of the integral will make the result positive, as it should be when you move against the direction of the electric field.

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