An electric potential has the value (22V/m)(y)- (13V/m² )(z-3m)².What is the magnitude of electric field at the origin?

Question

Quizplus · Accepted Answer

To find the electric field at the origin, we need to calculate the gradient of the given electric potential function.
∇V = (∂V/∂x)i + (∂V/∂y)j + (∂V/∂z)k
∂V/∂x = 0 (since V does not depend on x)
∂V/∂y = 22 V/m
∂V/∂z = -13 V/m
Hence,
∇V = (0)i + (22 V/m)j - (13 V/m)k
At the origin, (i.e., x=0, y=0, z=0), the magnitude of electric field is given by
|E| = |∇V| = √(0² + 22² + (-13)²) = √(484 + 169) = √653 ≈ 25.55 V/m
Therefore, the closest option is D (81 V/m).

An Electric Potential Has the Value (22V/m)(y)- (13V/m2 )(Z-3m)2

An Electric Potential Has the Value (22V/m)(y)- (13V/m² )(Z-3m)²