The electric field in a region is given by $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}}\to
 {\vec E} = \left( {Ax + B} \right)\hat i$, where E is in NC-1 and x is in metres. The values of constants are A=20 SI unit and B=10 SI unit. If the potential at x=1 is ${V_2}$ and that at x=−5 is ${V_2}$, then ${V_1} - {V_2}$ is
A) -48V
B) -520V
C) 180V
D) 320V

Hint: Applying electrified formulas use the given thing , applying integration for both sides there we easily get ${V_1} - {V_2}$ value by putting the values of A and B. In the electrodynamic chapter we have seen questions like this. In this question we ask the potential difference its unit is V.

Formula used:
$dV = - \vec E.d\vec r$

Complete Step by step solution:
$dV = - \vec E.d\vec r$

First we have to write given values
$\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}}\to
 {\vec E} = \left( {Ax + B} \right)\hat i$

Constant values are A=20 SI unit and B=10 SI unit. If the potential at x=1 is ${V_2}$ and that at x=−5 is ${V_2}$ , then ${V_1} - {V_2}$ is
$\eqalign{
  & \int\limits_{{v_2}}^{{v_1}} {dV = \int\limits_{ - 5}^1 { - \left( {Ax + B} \right)dx} } \cr
  & {v_1} - {v_2} = {\left( { - A\dfrac{{{x^2}}}{2} - Bx} \right)^1}_{ - 5} \cr} $

By simplification we get
$\eqalign{
  & \Rightarrow {v_1} - {v_2} = \left( { - \dfrac{A}{2} - B} \right) + \left( {\dfrac{A}{2}25 + B( - 5)} \right) \cr
  & \Rightarrow {v_1} - {v_2} = 12A - 6B \cr
  & \Rightarrow {v_1} - {v_2} = 240 - 60 \cr
  & \therefore {v_1} - {v_2} = 180V \cr} $

Hence, the correct option C.

Additional information:
The negative sign of $dV = - \vec E.d\vec r$this formula shows the direction of E, it is in the direction in which V decreases.

In other words we can say electric field points in the opposite direction to the voltage drop. And it is quite obvious because as you are going in the direction of the electric field you are in a way moving towards negative charges so electric potential is bound to decrease.

We have to practice the derivation of this formula. Then we get a strong hold on to use this formula.

Here we dr is the component of dr in the direction $\hat i$ dr can be either positive or negative depending on which way the external force is displaced. Once this expression becomes an integral then the sign of dr is determined by the limits of integration as we can see here.

Note:
If intensity of electric field is non uniform with respect to distance 'r' , over a given space, we have to use this formula for small segments in which considering uniformity of electric field intensity in that segment. And here we have to remember the negative sign of this formula.