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The electric potential V at any point (x, y, z), all in meters, in space is given by $V = 4x^2$ volt. The electric field at the point (1, 0, 2) in volt/meter is
A
16 along positive X-axis
B
8 along negative X-axis
C
8 along positive X-axis
D
16 along negative X-axis
Detailed Solution
The electric field is the negative gradient of the potential: $E_x = -\frac{\partial V}{\partial x}$, $E_y = -\frac{\partial V}{\partial y}$, $E_z = -\frac{\partial V}{\partial z}$
$V = 4x^2$ depends only on x, so $E_y = E_z = 0$.
$E_x = -\frac{\partial}{\partial x}(4x^2) = -8x$
At the point (1, 0, 2), x = 1: $E_x = -8$ V/m
The negative sign shows the field is along the negative X-axis.
So the field is 8 V/m along the negative X-axis.
$V = 4x^2$ depends only on x, so $E_y = E_z = 0$.
$E_x = -\frac{\partial}{\partial x}(4x^2) = -8x$
At the point (1, 0, 2), x = 1: $E_x = -8$ V/m
The negative sign shows the field is along the negative X-axis.
So the field is 8 V/m along the negative X-axis.