MCQ
The electric potential decreases uniformly from $120V$ to $80V$ as one moves on the $x-$axis from $x = -1\ cm$ to $x = + 1\ cm.$ The electric field at the origin.
  • Must be equal to $20Vcm^{-1}$
  • B
    May be equal to $20Vcm^{-1}$
  • C
    May be greater than $20Vcm^{-1}$
  • D
    May be lees than $20Vcm^{-1}$

Answer

Correct option: A.
Must be equal to $20Vcm^{-1}$

$\triangle\text{V}=-\text{E}.\text{dr}$
$(\text{V}_\text{f}-\text{V}_\text{i})=-\text{E}.\text{dr}=\text{E}_\text{x}(\text{B}-\text{A})$
$(80-120)=-\text{E}_\text{x}.(2)$
$\text{E}_\text{x}=\frac{40}{2}=20\text{v/m}$
If electric field lines lies in $'x\ '$ direction than it may be equal to $20v/cm.$
If Electric field lines liesin $'x - y'$ direction than it may be greater than $20v/cm.$

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