Electric potential V is given as a function of distance x (metre) by V = (5{x^2} + 10x - 9),volt . V

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2. Electric Potential and Capacitance

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The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5{x^2} + 10x - 9)\,volt$. Value of electric field at $x = 1$ is......$V/m$

A

$-20$

B

$6$

C

$11$

D

$ - 23$

Solution

(a) $E = – \frac{{dV}}{{dx}} = – \frac{d}{{dx}}(5{x^2} + 10x – 9) = – 10x – 10$
${(E)_{x = 1}} = – 10 \times 1 – 10 = – \,20\,V/m$

Standard 12

Physics

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