Why is $\Delta V = - \int E \cdot dr$ when charge is moving from lower potential to higher potential? 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.
 A: The expression
$$
\Delta U_F = -W_F
$$
is the definition of the potential energy function for a conservative force; no assumptions are being made here.  Then the definition of the change in potential is
$$
\Delta V = \frac{\Delta U_{F_E\text{ on }q_{\rm test}}}{q_{\rm test}}
$$
So, to find the potential difference in an electric field, you must determine the potential energy difference experienced by a test charge, which, in turn, is found by the work done by the electric force on that test charge.  In summary:
\begin{align}
\Delta V \left(\vec{r}_1 \rightarrow \vec{r}_2\right) &= \frac{\Delta U_{F_E \text{ on } q_{\rm test}} \left(\vec{r}_1 \rightarrow \vec{r}_2\right)}{q_{\rm test}}\\
 & = \frac{- W_{F_E \text{ on } q_{\rm test}} \left(\vec{r}_1 \rightarrow \vec{r}_2\right)}{q_{\rm test}}\\
 & = - \frac{1}{q_{\rm test}} \int_{\vec{r}_1 \rightarrow \vec{r}_2} \vec{F}_{E}\left(\vec{r}\right) \cdot d\vec{r}\\
& = - \frac{1}{q_{\rm test}} \int_{\vec{r}_1 \rightarrow \vec{r}_2} q_{\rm test} \, \vec{E}\left(\vec{r}\right) \cdot d\vec{r}\\
& = -  \int_{\vec{r}_1 \rightarrow \vec{r}_2} \vec{E}\left(\vec{r}\right) \cdot d\vec{r}\\
& = - \int_{t_1}^{t_2} \vec{E}\left(\vec{r}(t)\right) \cdot \frac{{\rm d} \vec{r}}{{\rm d} t}(t) \,dt\\
\end{align}
where the last expression is the meaning of the line integral, with $\vec{r}(t)$ a chosen parameterized path along the curve from $\vec{r}_1$ to $\vec{r}_2$. So the expression you wrote down for $\Delta V$ is completely general.
For the specific case of two (infinite) parallel-plate capacitors separated by $d$, the electric field is uniform, and we can choose a straight path, perpendicular to the plates. If you want to find $\Delta V_{\text{neg to pos}}$, then along this path $\vec{E}\cdot\vec{v}$ is negative, so:
\begin{align}
\Delta V &= - \int_{t_1}^{t_2} \vec{E}\left(\vec{r}(t)\right) \cdot \frac{{\rm d} \vec{r}}{{\rm d} t}(t) dt\\
&= + E \int_{t_1}^{t_2} \frac{{\rm d} x}{{\rm d} t}(t) \, dt\\
&= E \int_{x_{\rm neg}}^{x_{\rm pos}} {\rm d} x \\
&= E d
\end{align}
The result is positive, as it should be because the positive plate, $V_f$, is at a higher potential than the negative plate, $V_i$.
If you decided to calculate $\Delta V_{\text{pos to neg}}$, you would find a negative answer.
A: 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.
A: 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 $C$.
