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2. Electric Potential and Capacitance
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Electric potential at any point is $V = - 5x + 3y + \sqrt {15} z$, then the magnitude of the electric field is
A
$3\sqrt 2 $
B
$4\sqrt 2 $
C
$5\sqrt 2 $
D
$7$
Solution
(d) ${E_x} = – \frac{{dV}}{{dx}} = – ( – 5) = 5;$${E_y} = – \frac{{dV}}{{dy}} = – 3$
and ${E_z} = – \frac{{dV}}{{dz}} = – \sqrt {15} $
${E_{net}} = \sqrt {E_x^2 + E_y^2 + E_z^2} = \sqrt {{{(5)}^2} + {{( – 3)}^2} + {{( – \sqrt {15} )}^2}} = 7$
Standard 12
Physics
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