Question

in a certain region of space the potential is given by v is equal to K 2 x square minus y square + Z Square the electric field at the point 111 has magnitude​

Answer 1

use partial differential to determine the components of the electric field strength vector.

the electric field strength vector is given by

E = -k(4xi^ - 2yj^ + 2zk^)

therefore at the point (1, 1, 1) the magnitude of the electric field strength is (24)(1/2). HERE YOUR ANSWER BUDDY

Answer 2

Through the definition of electric potential, we can conclude that:

$|E_{x}| = |\frac{\delta V}{\delta x}|$

(yes, partial differentiation is used here)

For partially differentiating the given potential function with respect to x, treat all other variables (y, z) as constants.

So, the differentiation is:

$\frac{\delta V}{\delta x} = K(4x + 0 + 0)$

Similarly, the partial differentiation of the potential function w.r.t. y and z give the magnitudes of the electric field along y and z axes respectively, which are:

$\frac{\delta V}{\delta y} = K(2y)$

$\frac{\delta V}{\delta z} = K(2z)$

So:

$|E_{x}| = |K|4x, |E_{y}| = |K|2y, |E_{z}| = |K|2z$

The net magnitude of the electric field shall be:

$|\vec{E}| = \sqrt{E_{x}^2 + E_{y}^2 + E_{z}^2}$

So: $|\vec{E}| = \sqrt{K^2(16x^2 + 4y^2 + 4z^2)} = 2|K|\sqrt{4x^2 + y^2 + z^2}$

Finally, for the point (1, 1, 1), the field comes out to be:

$|\vec{E_{(1, 1, 1)}}| = 2|K|\sqrt{4 + 2 + 2} = 2|K|\sqrt{8} = 4\sqrt{2}|K|$