If the electric potential is given by V = (-1 volts/m4 )x2yz + (4 volts/m2)yz, find the electric field.
E= - gradφ=
= - {(δφ/δx)i̅+(δφ/δx)j̅ + (δφ/δx)k̅}=
= 2yz• i̅ –(4z-2xz)• j̅–(4y-2xy)•k̅.
E= - gradφ=
= - {(δφ/δx)i̅+(δφ/δy)j̅ + (δφ/δz)k̅}=
= 2yz• i̅ –(4z-2xz)• j̅–(4y-2xy)•k̅.
To find the electric field, we need to take the gradient of the electric potential function V. The gradient is a vector operator that gives the direction and magnitude of the steepest increase of a scalar field.
The electric field, E, can be found by taking the negative gradient of the electric potential, V:
E = -∇V
where ∇ is the del operator (also called nabla), which represents the gradient.
Now let's calculate the electric field using the given electric potential:
Given: V = (-1 volts/m^4)x^2yz + (4 volts/m^2)yz
To calculate the gradient, we differentiate the potential with respect to each variable (x, y, z) and multiply by their respective unit vectors (i, j, k):
∇V = (∂V/∂x)i + (∂V/∂y)j + (∂V/∂z)k
Taking the partial derivatives:
∂V/∂x = -2(-1 volts/m^4)x(yz) = 2x(yz) volts/m^4
∂V/∂y = x^2z(-1 volts/m^4) + 4z volts/m^2 = -x^2z volts/m^4 + 4z volts/m^2
∂V/∂z = x^2y(-1 volts/m^4) + 4y volts/m^2 = -x^2y volts/m^4 + 4y volts/m^2
Putting it all together:
∇V = (2x(yz))i + (-x^2z volts/m^4 + 4z volts/m^2)j + (-x^2y volts/m^4 + 4y volts/m^2)k
So, the electric field E is given by:
E = -∇V = - (2x(yz))i - (-x^2z volts/m^4 + 4z volts/m^2)j - (-x^2y volts/m^4 + 4y volts/m^2)k
Simplifying this expression will give you the electric field vector.