91Ó°ÊÓ

Electric potential is the measure of the potential energy per unit charge at a specific point in an electric field. It is represented by the symbol \(V\) and is measured in volts. Imagine it like a hill where charges 'roll'. The higher you are on the hill, the more potential energy you have. In this exercise, the electric potential is defined as \(V = 2.00xyz^2\).
This tells us how the potential changes with variations in \(x\), \(y\), and \(z\). Each of these variables plays a role in determining the energy landscape, like crafting the slopes and ridges on our 'hill'.

• x, y, z represent position in space, and each has an effect on potential.
• The equation shows potential increasing with \(xy\) and particularly powerful with \(z^2\).

Understanding and calculating electric potential is crucial for grasping how electric fields influence charged particles.

The gradient of a field points in the direction of the largest space-related increase of that field. Think of it as the steepest path on our energy 'hill'. The gradient helps us understand how rapidly and in what direction the potential differs.
In mathematical terms, for the potential \(V(x, y, z)\), the gradient is represented as \(abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right)\). This captures the rate of change across each axis:

• \(\frac{\partial V}{\partial x}\): Rate of change of potential along the x-axis.
• \(\frac{\partial V}{\partial y}\): Rate of change along the y-axis.
• \(\frac{\partial V}{\partial z}\): Effect along the z-axis.

The gradient serves as a bridge to finding the electric field from the potential.

Partial derivatives illustrate how a function changes with respect to one variable alone, holding others constant. This is like examining how our 'hill' steeply rises or falls when moving solely in one direction, unaffected by changes elsewhere.
For the potential \(V = 2.00xyz^2\), the partial derivatives are:

• \(\frac{\partial V}{\partial x} = 2.00yz^2\).
• \(\frac{\partial V}{\partial y} = 2.00xz^2\).
• \(\frac{\partial V}{\partial z} = 4.00xyz\).

These expressions show how potential changes along each axis. Mastering the concept of partial derivatives helps in understanding gradients and ultimately the electric field dynamics.
Knowing these values at a certain point, which we calculate in the exercise, is crucial for practical applications in physics.

The magnitude of the electric field tells you about the strength of the field at a given point. This is similar to measuring how fast and powerfully an object might be pushed off our 'hill'. The electric field vector \(\mathbf{E}\) can be derived from the negative gradient of the potential: \(\mathbf{E} = -abla V\).
From the gradient, substitute our calculated partial derivatives and negate them:

• At our given point, \(abla V\) resulted in \( (64.00, -96.00, 96.00) \).
• Then electric field \(\mathbf{E} = (64.00, -96.00, 96.00)\).

The magnitude of this vector is calculated using Pythagoras' theorem:\(|\mathbf{E}| = \sqrt{64.00^2 + (-96.00)^2 + 96.00^2} \approx 150.09 \text{ V/m}\).
This magnitude provides insight into the field's intensity, fundamental for predicting how charged particles move within the field.