An electric field is surrounding an electric charge and also exerting force on other charges in the field at the same time. It is either attracting or repelling them. The electric field is many times abbreviated as E-field. The electric field is defined mathematically like a vector field that associates to each point in the space the (electrostatic or Coulomb) force/unit of charge exerted on an infinitesimal positive test charge is at rest at that particular point. The SI unit of electric field strength is volt/meter. (V/m). Newtons/coulomb (N/C) is also useful as a unit of electric field strength. Electric fields are generated by electric charges, or by the time-varying magnetic fields. Learn the electric field formula here.

**Formula and Derivation of Electric Field**

E = -\(\frac{\Delta v}{d}\) is derived from the definition of the electric potential difference \(\Delta V\). Pay decent attention to the negative signs in the derivation, as that is the source of many errors.

Briefly, we imagine some distribution of fixed charged particles that are producing a net electric field \(\vec{E}\). Then we can calculate the work W that we need to move a rest charge at a constant velocity from point a to point b along some arbitrary path:

W = \(\int_{a}^{b}\vec{F}\cdot d\vec{s}\),

where \(\vec{F}\) is the force that is applied against the net force \(\vec{F}_{E}\) because of the electric field, and d\(\vec{s}\) is a differential displacement vector along the way from a to b. We want the rest particle to move with a constant velocity, So, \(\vec{F} = -\vec{F}_{E}\). Therefore, the work done by the electric field here will be:

W = -\(\int_{a}^{b}\vec{F}_{E}\cdot d\vec{s}\).

Next, we are considering the work done by the electric field/unit charge:

\(\frac{W}{qunit}\) = \(\Delta\) V = -\(\int_{a}^{b}\vec{E}\cdot d\vec{s}\).

The integral never depends on the path between a and b, not just because of something to do with this kind of integral, but because we have an assumption that the source charges that generate \(\vec{E}\) are fixed so that \(\vec{E}\) don’t change as we are moving the charge at rest. This clears that for fixed \(\vec{E}\), V depends only on a and b, which are just points in space.

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The next part needs us to draw a figure: a closed curve of any kind of shape with points i.e. a and b on the curve. Mark a third point c on the same curve. The work done moving the test charge from point c to point a will be here: \(W_{c\rightarrow a}\), and from point c to point b is \(W_{c\rightarrow b}\), so that:

W = -\(\int_{a}^{b}\) \(\vec{F}_{E}\cdot d\vec{s}\) = \(W_{c\rightarrow b}\) – \(W_{c\rightarrow a}\)

$$ W_{c\rightarrow c} = 0 = W_{c\rightarrow a} + W_{a\rightarrow b} + W_{b\rightarrow c} \rightarrow W_{a\rightarrow b} = -W_{b\rightarrow c} – W_{c\rightarrow a}$$

But $$W_{b\rightarrow c} = -W_{c\rightarrow b}$$

so $$W_{a\rightarrow b} = W_{c\rightarrow b} – W_{c\rightarrow a}$$

Now, \(W_{c\rightarrow b}\) and \(W_{c\rightarrow a}\) are scalars and whose values are depending upon on the choice of point c. So choosing point c is equivalent to choosing any scalar field V whose value at point c can be set to 0, and we might write:

\(\frac{W}{qunit}\) = V\(\left ( b \right )-V\left ( a \right )\).

It follows:

\(\int_{a}^{b}\vec{E}\cdot d\vec{s}\) = V\(\left ( a \right )\) – V\(\left ( b \right )\) = -\(\Delta V\).

As a specific case, Let’s consider a uniform electric field \(\vec{E}\). The integral becomes E\(\Delta\) s, where \(\Delta\) s is the displacement in the direction of E. If we let \(\Delta\) s =d, we have:

E = -\(\frac{\Delta V}{d}\).

## Solved Example Electric Field Formula

#### Question:

A charge of q = – 4.0 × \(10^{-6}\) is placed in an electric field and experiences a force of 5.5 N [E]. If charge q is removed, what is the magnitude and direction of the force exerted on a charge of – 2q at the same location as charge q?

#### Solution:

The force on a charge -2q due to an electric field E is given by

F2 = -2 q E = -2(q E) = -2(5.5[E]) = -11 [E] or 11 N [W].