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So in a region of free space with zero charge, all I'm given is that the electrostatic potential V is given by:

$$V(x,y,z) = kz(x^2) y + 2y - 6y(z^3)$$ [Volts]

To find the constant k, would I have to use the assumption that there is zero electric field and that the corresponding partial derivatives dV/dx, dV/dy, and dV/dz also equal to zero?

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    \$\begingroup\$ Between the plates of a charged capacitor in a vacuum there is no charge. And yet the field is not zero. You should try to catch another fish, mon ami. \$\endgroup\$ – Sredni Vashtar Mar 1 at 1:18
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Why electric field is zero? It is a gradient of potential

$$ E=-grad(V) $$

Now use Maxwell's equation for electric field taking into account that there is now electric charge in your space

$$ \mathrm{div}\,E = 0 $$

Then combine the two equations

$$ \mathrm{div}\,\mathrm{grad}\, V=\Delta V=0 $$

Where \$\Delta\$ is a Laplace operator.

Apply the operator to your equation for V and you should be able to solve it for k.

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