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The electric potential V is given by as a function of distance \mathrm{ x (metre) by \, \, \, v=\left(5 x^2-10 x-9\right) Volts }. The value of electric field at x=1 \mathrm{\mathrm{~m} } is -

Option: 1

20 \mathrm{v} / \mathrm{m}


Option: 2

6 \mathrm{v} / \mathrm{m}


Option: 3

11 \mathrm{v} / \mathrm{m}


Option: 4

zero


Answers (1)

best_answer

Given,
\mathrm{\begin{aligned} & v=5 x^2-10 x-9 \\ & E=-\frac{d v}{d x} \\ & E=-\frac{d}{d x}\left[5 x^2-10 x-9\right] \end{aligned}}
\mathrm{ \vec{E}=-[10 x-10] Electric\, \, field\, at\, x=1 \mathrm{~m} }
\mathrm{ \begin{aligned} & \vec{E}=-[10 \times 1-10]=0 . \\ & \vec{E}=0 \mathrm{~V} / \mathrm{m} \text { Ans } \end{aligned} Ans }
 

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avinash.dongre

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