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Physics Question on electrostatic potential and capacitance

If potential (in volts) in a region is expressed as V(x,y,z)V(x, y , z) = 6xy−y+2yz6xy - y + 2yz, the electric field (in N/C)(in\, N/C) at point (1,1,0)(1, 1, 0) is

A

−(2i^+3j^+k^) -(2\widehat {i} + 3\widehat {j} + \widehat {k} )

B

−(6i^+9j^+k^) -(6\widehat {i} + 9\widehat {j} + \widehat {k} )

C

−(3i^+5j^+3k^) -(3\widehat {i} + 5\widehat {j} + 3\widehat {k} )

D

−(6i^+5j^+2k^) -(6\widehat {i} + 5\widehat {j} + 2\widehat {k} )

Answer

−(6i^+5j^+2k^) -(6\widehat {i} + 5\widehat {j} + 2\widehat {k} )

Explanation

Solution

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