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Engineering LibreTexts

2.2: Electric Field Intensity

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  • Electric field intensity is a vector field we assign the symbol \(\mathbf{E}\) and has units of electrical potential per distance; in SI units, volts per meter (V/m). Before offering a formal definition, it is useful to consider the broader concept of the electric field.

    Imagine that the universe is empty except for a single particle of positive charge. Next, imagine that a second positively-charged particle appears; the situation is now as shown in Figure 2.1. Since like charges repel, the second particle will be repelled by the first particle and vice versa. Specifically, the first particle is exerting force on the second particle. If the second particle is free to move, it will do so; this is an expression of kinetic energy. If the second particle is somehow held in place, we say the second particle possesses an equal amount of potential energy. This potential energy is no less “real,” since we can convert it to kinetic energy simply by releasing the particle, thereby allowing it to move

    Now let us revisit the original one particle scenario.


    Figure 2.1: A positively-charged particle experiences a repulsive force \(\mathbf{F}\) in the presence of another particle which is also positively-charged.


    Figure 2.2: A map of the force that would be experienced by a second particle having a positive charge. Here, the magnitude and direction of the force is indicated by the size and direction of the arrow.

    In that scenario, we could make a map in which every position in space is assigned a vector that describes the force that a particle having a specified charge \(q\) would experience if it were to appear there. The result looks something like Figure 2.2. This map of force vectors is essentially a description of the electric field associated with the first particle.

    There are many ways in which the electric field may be quantified. Electric field intensity \(\mathbf{E}\) is simply one of these ways. We define \(\mathbf{E(r)}\) to be the force \(\mathbf{F(r)}\) experienced by a test particle having charge \(q\), divided by q; i.e.,

    \[\mathbf { E } ( \mathbf { r } ) \triangleq \lim _ { q \rightarrow 0 } \frac { \mathbf { F } ( \mathbf { r } ) } { q }\]

    Note that it is required for the charge to become vanishingly small (as indicated by taking the limit) in order for this definition to work. This is because the source of the electric field is charge, so the test particle contributes to the total electric field. To accurately measure the field of interest, the test charge must be small enough not to significantly perturb the field. This makes Equation 2.1 awkward from an engineering perspective, and we’ll address that later in this section.

    According the definition of Equation 2.1, the units of E are those of force divided by charge. The SI units for force and charge are the newton (N) and coulomb (C) respectively, so E has units of N/C. However, we typically express E in units of V/m, not N/C. What’s going on? The short answer is that 1 V/m = 1 N/C: