Find electric potential due to charge distribution

[David23454]

Homework Statement

In a certain region, the electric potential due to a charge distribution
is given by the equation V (x, y, z) = (3x²y²+yz³-2z³x)V₀/a⁴ where
a, x, y, and z are measured in meters and V and V₀ are in volts. What
is the magnitude of the electric field at the position (x, y, z) = (a, a, a)?

Homework Equations

[/B]
E=-(dV/dx(i)+dV/dy(j)+dV/dz(k))

The Attempt at a Solution

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Taking the negative derivative of V(x,y,z) and inputing "a" gives E=-V₀/a(4,7,-3)

I would have thought that this was the complete solution, but the solution that goes on to absval(E)=V₀/a(4²+7²+3²)^½
=V₀/a(74)^½

I haven't taken multi-variable calculus yet (it wasn't a requirement for the course), so I'm a little confused as to what's going on in the part where the absolute value of E is equal to V₀/a4(²+7²+3²)^½. Could someone explain to me how this works, specifically, why should I take the abs value of E, and add up the squares of the derivatives and then take their square root to get the answer? Thanks!

 

[Henryk]

You are almost there.
Electric potential is a scalar. Electric field is a vector. The formula you are using is correct.The x-component of the electric field is $-\frac {dV}{dx}$, etc.
You applied the formula correctly and your answer (in a vector form) is correct. The problem asks you to find the magnitude of the field, that is the length of the vector. Yes, you square all the components, sum them up and take the square root.