Question

In a certain region of space, the potential is given by v=2x−5x2y 3yz2. How much is the magnitude of the electric field at point (-2, 2, 0)?

Answer 1

The magnitude of the electric field at the point (-2, 2, 0) is 46.52 V/m

How to calculate the electric field?

Since in a certain region of space, the potential is given by v = 2x − 5x²y 3yz²,

The electric field, E = -grad(V) where grad(V) = (dV/dx)i + (dV/dy)j + (dV/dz)k

Since V = 2x − 5x²y + 3yz²

dV/dx = d(2x − 5x²y + 3yz²)/dx

= d2x/dx - d5x²y/dx + d3yz²/dx

= 2 - 10xy + 0

= 2 - 10xy

dV/dy = d(2x − 5x²y + 3yz²)/dy

= d2x/dy - d5x²y/dy + d3yz²/dy

= 0 - 5x² + 3z²

=  - 5x² + 3z²

dV/dz = d(2x − 5x²y + 3yz²)/dz

= d2x/dz - d5x²y/dz + d3yz²/dz

= 0 - 0 + 6z

=  6z

The electric field

So, E = -grad(V)

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

= -[(2 - 10xy)i + (- 5x² + 3z²)j + (6z)k]

So, the electric field at (-2, 2, 0) is

E = -[(2 - 10xy)i + (- 5x² + 3z²)j + (6z)k]

E = -[(2 - 10(-2)(2))i + (- 5(2)² + 3(0)²)j + (6(0))k]

E = -[(2 + 40)i + (- 5(4) + 0)j + (0)k]

E = -[42i + (-20 + 0)j + (0)k]

E = -[42i - 20j + 0k]

E = -42i + 20j - 0k

The magnitude of the electric field

So, the magnitude of E at (-2, 2, 0) is

|E| = √[(-42)² + 20² + 0²]

=  √[1764 + 400 + 0]

= √2164

= 46.52 V/m

So, the magnitude of the electric field at the point (-2, 2, 0) is 46.52 V/m

Learn more about electric field here:

https://brainly.com/question/25751825

#SPJ12