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# 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 ${V_1} - {V_2}$ isA) -48VB) -520VC) 180VD) 320V

Last updated date: 20th Jun 2024
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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.

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.
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.