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