At any point (x, 0, 0), the electric potential V is 10x+15x2+5x3 volt, then electric field at x = 1 m is

E→=−dvdxi^=−dVdx10x+15x2+5x3i^=−−10x2−30x3−15x4i^At x = 1E→=10+30+15=55i^

At any point (x, 0, 0), the electric potential V is 10x+15x2+5x3 volt, then electric field at x = 1 m is

E→=−dvdxi^=−dVdx10x+15x2+5x3i^=−−10x2−30x3−15x4i^At x = 1E→=10+30+15=55i^

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At any point (x, 0, 0), the electric potential V is $(\frac{10}{x} + \frac{15}{x^{2}} + \frac{5}{x^{3}}) volt,$ then electric field at x = 1 m is

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$55 (\hat{j} + \hat{k}) \lor / m$

$55 \hat{iv} / m$

$\frac{55}{\sqrt{2}} (\hat{i} + \hat{k}) v / m$

$55 \hat{j} v / m$

answer is B.

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Detailed Solution

$\vec{E} = - \frac{dv}{dx} \hat{i}$

$= - \frac{dV}{dx} [\frac{10}{x} + \frac{15}{x^{2}} + \frac{5}{x^{3}}] \hat{i}$

$= - [- \frac{10}{x^{2}} - \frac{30}{x^{3}} - \frac{15}{x^{4}}] \hat{i}$

At x = 1

$\vec{E} = 10 + 30 + 15$

$= 55 \hat{i}$

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At any point (x, 0, 0), the electric potential V is 10x+15x2+5x3 volt, then electric field at x = 1 m is

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At any point (x, 0, 0), the electric potential V is $(\frac{10}{x} + \frac{15}{x^{2}} + \frac{5}{x^{3}}) volt,$ then electric field at x = 1 m is