Systems of Charged Objects

This Page belongs to Laura Breithaupt

In many cases in the real world, there are more than one charge contributing to an interaction. However, in some simple cases, one can choose a system (and ignore external forces) by choosing a system of charges and observing the changes of Kinetic energy (K_sys) and Potential energy (U).

The Main Idea

In physics, a physical system is a region of the physical universe to be analyzed, and everything outside the system is known as the "surroundings." The same logic can be applied to a system of charges; only energy changes within the system need to be analyzed.

A Mathematical Model

ΔK_sys + ΔU = 0 (zero work done by surroundings)

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Examples

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Simple

Consider a system of two uniformly charged plates of a capacitor and a single proton. In the region between the plates, there is nearly uniform electric field and external forces can be ignored. The proton is traveling through the right between the plates.

While the proton is between the plates, an electric force is acting on it. The direction of the force is to the right (Electric field points to the right: away from the positive plate and toward the negative plate) F_electric = Q_proton * E

Question: Will the Kinetic Energy (K) of the proton increase or decrease as a result of this force?

Answer: Because the force is in the same direction as the motion of the particle (the displacement, Δx), the Kinetic Energy will increase.

ΔK_proton + ΔU_electric = 0

The change in electric potential energy is equal to the negative of internal work of the system.

ΔU_electric = -(W_intenal) = -(F_electric * Δx) ΔU_electric = -(Q_proton * E * Δx)

Because E and Δx are in the same direction, their product is positive.

ΔK_proton + ΔU_electric = 0 ΔK_proton = -(ΔU_electric) ΔK_proton = Q_proton * E * Δx , which is a positive value

Kinetic energy of the proton increases because ΔK_proton is positive.

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