Electric field strength and potential gradient

[zinc79]

A bit of a problem. My book teaches me that E = -(dV/dx), where E is the electric field strength, V is the electric potential, and x represents displacement.

But, it also suggests along with the above formula that E = -(V/d) and displays a circuit with a battery of p.d. V and two parallel metal plates of distance (d) from each other.

My question is, HOW did E = -(dV/dx) become E = -(V/d)? The former formula, proven via differentiation, says that the electric field strength is negative of the potential gradient i.e. rate of change of electric potential with respect to the displacement. Then how does this transform into the electric field strength simply being equal to the negative of the ratio of the potential difference to the distance? It makes no sense to me!

And yet, I've seen the latter formula being used in practice questions.

 

[Gear300]

If the rate of change of V is constant over the distance x (or d), then -dV/dx can be stated as -V/d. -V/d is not equal to -dV/dx, but in a parallel plate capacitor it usually is.

 

[robphy]

To follow up on Gear300's comment...

the general relation is E = -(dV/dx)...
for parallel plates, one really should write E_{parallel plates} = -(V/d)

Recall for parallel plates, E_{parallel plates} is a constant [between the plates].

(Don't merely match symbols or abbreviations.
Not all "E"s mean the same thing.
For each "formula", understand what the variable represents.)

 

[zinc79]

Hey, thank you for replying.

See, what I forgot to mention was my perspective on the parallel plate capacitor, which yes, has a uniform electric field. There is NO change in the p.d. as we move from one plate to the other, so shouldn't dV/dx become zero?

 

[ZapperZ]

Er.. Unless I misread the OP, the confusion here comes from the fact that these are NOT the same "V".

In E = -dV/dx, this V is the electrostatic potential at a position x, i.e. V(x). It has a particular value depending on geometry.

In E = V/d, this "V", really is the potential difference between the two plates. If I call this V' instead of V, then here you can equate V' to dV in the first equation, and d to dx.

So it is important to know what those symbols really mean, rather than simply looking at the symbol. Often the same symbol represents different things.

Zz.

 

[Domenicaccio]

Exactly, the first equation is about the electrical field E(x) and potential V(x) at each point in space along the direction x, while the second is a macroscopic relationship between an allegedly constant field E and a voltage V which is a quantity defined for the whole system.

The first one is the general relationship between electric field and potential, and is valid in any scenario.

The second one is a particular case of the general relationship. You can obtain it by assuming E(x)=E constant and integrating in dx.

 

[jtbell]

The parallel-plate equation $E = -V/d$ really should read $E = -\Delta V/d$ to indicate that the voltage is the difference in potential between the two plates. Unfortunately it's a long-standing custom to use V to represent both potential (at a point) and potential difference (between two points) in different contexts.

 

[granpa]

The parallel-plate equation $E = -V/d$ really should read $E = -\Delta V/d$ to indicate that the voltage is the difference in potential between the two plates. Unfortunately it's a long-standing custom to use V to represent both potential (at a point) and potential difference (between two points) in different contexts.

good answer.