Electric Charges and Fields - Result Question 19

19. The electric potential $V$ at any point $(x, y, z)$, all in meters in space is given by $V=4 x^{2}$ volt. The electric field at the point $(1,0,2)$ in volt/ meter is

(a) 8 along positive $X$-axis

[2011M]

(b) 16 along negative $X$-axis

(c) 16 along positive $X$-axis (d) 8 along negative $X$-axis

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Answer:

Correct Answer: 19. (d)

Solution:

(d) $\overrightarrow{{}E}=-[\frac{d V}{d x} \hat{i}+\frac{d V}{d y} \hat{j}+\frac{d V}{d z} \hat{k}]$

Given, $V=4 x^{2}$

$\therefore \quad E=-i \frac{d(4 x^{2})}{d x}$

$=-8 x \hat{i}$ volt $/$ meter

$\therefore \overrightarrow{{}E} _{(1,0,2)}=-8 \hat{i} V / m$