1) The electric potential at any point x, y, z in metres is given by V = 3x2. The electric field at a point (2 m, 0, 1 m) is
To find the electric field at a specific point, we need to determine the gradient of the electric potential function.
The gradient is a vector operator that tells us the rate of change of a function in all directions. In Cartesian coordinates (x, y, z), the gradient is given by:
∇V = (∂V/∂x)i + (∂V/∂y)j + (∂V/∂z)k
Here, i, j, and k are unit vectors in the x, y, and z-directions, respectively.
In our case, the electric potential V = 3x^2, so let's calculate the partial derivatives:
∂V/∂x = 6x
∂V/∂y = 0 (since there is no y-dependence)
∂V/∂z = 0 (since there is no z-dependence)
Substituting these values into the gradient formula, we get:
∇V = (6x)i + 0j + 0k = 6xi
Now, we need to find the electric field E at the point (2 m, 0, 1 m). The electric field is related to the gradient of the electric potential by the equation:
E = -∇V
The negative sign is added because the electric field points from higher potential to lower potential. Therefore:
E = -(6xi) = -6xi
So, the electric field at the point (2 m, 0, 1 m) is -12i N/C. The direction of the electric field is negative x-direction.
To summarize the steps:
1. Calculate the partial derivatives of the electric potential with respect to x, y, and z.
2. Substitute the partial derivatives into the gradient formula to obtain the gradient.
3. Multiply the gradient by -1 to get the electric field vector.