# Find electric potential due to charge distribution

1. May 11, 2017

### David23454

1. The problem statement, all variables and given/known data

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)?

2. Relevant equations

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

3. The attempt at a solution

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!

2. May 11, 2017

### 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.