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Can someone help me with this, - Electrostatics - JEE Main-10

The electric filed in a region is given by \vec{E}=\left ( Ax+B \right )\hat{i}, where E is in NC^{-1} and x is in meters. The value of constantd are A=20 SI unit and B= 10 SI unit. If the potential at x=1 is V1 and that at x=-1 is V2 , then V1-V2 is:

 

  • Option 1)

    320 V

  • Option 2)

    -48 V

  • Option 3)

    180 V

  • Option 4)

    -520 V

 
Answers (1)
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Relation between E and V in integral form -

\dpi{100} dV=-\int_{r_{1}}^{r_{2}}\overrightarrow{E}\cdot \vec{d}r=-\int_{r_{1}}^{r_{2}}Edr\cos \theta

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V_{f}-V_{i}=-\int E.dx

V_{1}-V_{2}=\int_{-5}^{1}\left ( ax +b \right )dx

                =-\left ( \frac{ax^{2}}{2} +bx\right )_{-5}^{1}

               =-\left [\left ( \frac{a}{2} +b\right )-\left ( \frac{25a}{2} -5b\right ) \right ]

            =-\left [\left ( \frac{20}{2} +10\right )-250+50 \right ]

          =180V

             


Option 1)

320 V

Option 2)

-48 V

Option 3)

180 V

Option 4)

-520 V

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