91Ó°ÊÓ

In physics, differentiation is a powerful tool, especially when dealing with varying quantities such as electric potential. It's the process of finding the rate at which a quantity changes with respect to another. In the context of our exercise, it helps us determine how the electric potential changes with distance.
The electric potential, given by the equation \( V = 1500x^2 \), changes as you move away from plate 1. To find how this potential changes with distance \( x \), we differentiate it with respect to \( x \). This mathematical operation transforms the potential function into the electric field equation \( E = -\frac{dV}{dx} \).

• The differentiation of \( V = 1500x^2 \) with respect to \( x \) results in \( \frac{dV}{dx} = 3000x \). This denotes that at any point \( x \), the rate of change of the potential gives us the electric field.
• A negative sign in the electric field equation indicates that the field points in the direction of decreasing potential.

Thus, differentiation in physics serves as a bridge from potential functions to tangible, measurable fields.

The direction of the electric field reveals essential information about the forces acting on a charge placed in the field. It typically moves from regions of higher to lower potential. In the exercise, when determining the electric field's direction, we found that the electric field \( E \) was \(-39\, \text{V/m}\). This negative value isn't random; it tells us the direction of the field.
The negative sign in the calculated electric field value implies the field directs from higher potential towards lower potential. Specifically, in this scenario:

• Higher potential is at greater \( x \), further from plate 1.
• The negative sign of \( E = -39 \text{ V/m} \) indicates that the field points towards plate 1, since moving towards plate 1 decreases the potential.

Understanding the direction ensures that we can predict the movement of a positively charged particle within this field, as it would naturally move against the field direction, towards higher potential.

The magnitude of an electric field is a quantitative measure of the force experienced by a positive test charge placed in the field. In our specific case, we found the magnitude using the derivative of the potential function. After differentiating \( V = 1500x^2 \), we obtained \( E = 3000x \), and subsequently, \( E = -39 \, \text{V/m} \) at \( x = 0.013 \, \text{m} \).
Here's why this matters:

• The computed magnitude \( 39 \, \text{V/m} \) informs us how strong the electric field is at that point. It tells us the force per unit charge a positive test charge would feel there.
• The formula \( E = -\frac{dV}{dx} \) indicates a relationship where the strength of the field is directly tied to the rate of potential change with distance.

The magnitude, separate from direction, explicitly quantifies the field's strength, which is crucial for understanding the dynamics of charged particles in the vicinity of the field.