At any point the electrical potential is v=3x 2y² at (3,1) point. What will be the field at that point?

Question

At any point the electrical potential is v=3x   2y² at (3,1) point. What will be the field at that point?

Answer

Partial differentiate V wrt x and y

E=-( 3.2y2 i^+ 3x .4 j^)V/m
E=- ( 6y2 i^ + 12 x j^)

Answer

Calculation of Electric Field at a Given Point

Electric field (E) is given by the negative gradient of electric potential (V) at a point. Mathematically, E = -∇V.

Given Electric Potential Function

The given electric potential function is V = 3x - 2y².

Calculation of Partial Derivatives

To find the electric field at the point (3,1), we need to calculate the partial derivatives of the electric potential function with respect to x and y.

∂V/∂x = 3
∂V/∂y = -4y

Substitute the Given Point

At the point (3,1):
∂V/∂x = 3
∂V/∂y = -4

Calculate Electric Field Components

Now, we can calculate the electric field components:
Ex = -∂V/∂x = -3
Ey = -∂V/∂y = 4

Final Electric Field at the Given Point

Therefore, the electric field at the point (3,1) is E = -3i + 4j, where i and j represent the unit vectors in the x and y directions respectively.

This electric field vector represents the direction and magnitude of the electric field at the given point.

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