Derivation of formula for electric field strength [duplicate]

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I'm currently studying Cambridge A-Levels Based on the definition of electric field strength in the textbook, in which electric field strength at a point is the force per unit charge exerted on a charge at the point, the equation E = F/Q can be derived. However, there is another formula which can be derived: E = -V/d which I don't know how to derive. Any help would be greatly appreciated, Thanks !

marked as duplicate by sammy gerbil, Jon Custer, user191954, stafusa, ZeroTheHeroOct 6 '18 at 19:30

$$E = - \frac{\Delta V}{d}$$ is derived from the definition of electric potential difference $$\Delta V$$. Pay careful attention to the negative signs in the derivation, as that is probably the source of most errors by beginning students.

Briefly, we imagine some distribution of fixed charged particles that produce a net electric field $$\vec{E}$$. Then we calculate the work $$W$$ required to move a test charge at constant velocity from point $$a$$ to point $$b$$ along any arbitrary path:

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

where $$\vec{F}$$ is the force applied against the net force $$\vec{F}_{E}$$ due to the electric field, and $$d\vec{s}$$ is a differential displacement vector along the path from $$a$$ to $$b$$. We want the test particle to move with constant velocity, so $$\vec{F} = -\vec{F}_{E}$$. Therefore, the work done by the electric field is

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

Next, we consider the work done by the electric field per unit charge:

$$\frac{W}{q_{\textrm{unit}}} = \Delta V = -\int_{a}^{b}\vec{E}\cdot d\vec{s}.$$

The integral does not depend on the path between $$a$$ and $$b$$, not because of anything to do with this kind of integral, but because we have assumed the source charges that generate $$\vec{E}$$ are fixed so that $$\vec{E}$$ does not change as we move the test charge. This means that for fixed $$\vec{E}$$, $$V$$ depends only on $$a$$ and $$b$$, which are just points in space.

The next part requires you to draw a figure: a closed curve of any shape with points $$a$$ and $$b$$ on the curve. Mark a third point $$c$$ on the curve. The work done moving the test charge from point $$c$$ to point $$a$$ is $$W_{c \to a}$$, and from point $$c$$ to point $$b$$ is $$W_{c \to b}$$, so that

$$W = -\int_{a}^{b}\vec{F}_{E}\cdot d\vec{s} = W_{c \to b} - W_{c \to a}.$$

($$W_{c \to c} = 0 = W_{c \to a} + W_{a \to b} + W_{b \to c} \implies W_{a \to b} = -W_{b \to c} - W_{c \to a}$$. But $$W_{b \to c} = -W_{c \to b}$$, so $$W_{a \to b} = W_{c \to b} - W_{c \to a}$$.)

Now, $$W_{c \to b}$$ and $$W_{c \to a}$$ are just scalars whose values depend on the choice of point $$c$$. So choosing point $$c$$ is equivalent to choosing some scalar field $$V$$ whose value at point $$c$$ can be set to zero, and we may write

$$\frac{W}{q_{\textrm{unit}}} = V(b) - V(a).$$

It follows that

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

As a special case, 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 get

$$E = -\frac{\Delta V}{d}.$$