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 An electric field is given byE _x = - 2 x^3 kN/C. The potential of the point (1, –2), if potential of the point (2, 4) is taken as zero, is                                                      

Option: 1

- 7.5 \times 10 ^3 V


Option: 2

7.5 \times 10 ^3 V


Option: 3

- 15 \times 10 ^3 V


Option: 4

15 \times 10 ^3 V


Answers (1)

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As we learned

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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dV = - \bar E . d\bar r = - ( 2 x^3 \times 10 ^3 \hat i )(dx \hat i+ dy \hat j + dz \hat k ) = 2 x^3 \times 10 ^3 dx \\\\ \int_{0}^v{} dV = - \int_{2}^1{2x^3} \times 10 ^3 dx = V = 7.5 \times 10 ^3 V

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Anam Khan

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