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