Question

If potential (in volts) in a region is expressed as $V(x, y , z)$ = $6xy - y + 2yz$, the electric field $(in\, N/C)$ at point $(1, 1, 0)$ is

• A. $ -(2\widehat {i} + 3\widehat {j} + \widehat {k} )$
• B. $ -(6\widehat {i} + 9\widehat {j} + \widehat {k} )$
• C. $ -(3\widehat {i} + 5\widehat {j} + 3\widehat {k} )$
• D. $ -(6\widehat {i} + 5\widehat {j} + 2\widehat {k} )$

Answer 1

Answer: (D)

The electric field, $ \overrightarrow {E} $ and potential V in a region are related as
$ \overrightarrow {E} = - \bigg [ \frac{\partial V}{\partial x} \widehat {i} + \frac{\partial V}{\partial y} \widehat {j} + \frac{\partial V}{\partial z} \widehat {k} \bigg] $
Here, $V(x, y, z) = 6xy - y + 2yz$
$ \therefore \overrightarrow {E} = - \bigg [\frac{\partial}{\partial x} ( 6xy - y + 2yz) \widehat {i} + \frac{\partial}{\partial y} ( 6xy - y + 2yz) \widehat {j}$
$+ \frac{\partial}{\partial z} ( 6xy - y + 2yz) \widehat {k} \bigg] $
$ = -[(6(y)) \widehat{i} + (6x - 1 + 2z) \widehat {k} + (2(y)) \widehat {k}]$
At point (1, 1,0),
$ \overrightarrow {E} = -[(6(1)) \widehat{i} + (6(1) - 1 + 2(0)) \widehat {j} + (2(1)) \widehat {k}]$
$ = -(6 \widehat {i} + 5 \widehat {j} + 2 \widehat {k}) $