# E = V/d how?

Discussion in 'Physics' started by logearav, Dec 11, 2011.

1. ### logearav Thread Starter Member

Aug 19, 2011
248
0
Revered members,
I know dV = -E.dx so
E = -dV/dx
So E should be E = -V/x or E = -V/d
But how E = V/d?

2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
What equation are you working from?

The d shouldn't stand alone, it needs to indicate the variable that can be differentiated across the function range.

Generally, E=V

3. ### davebee Well-Known Member

Oct 22, 2008
539
46
I think logearav is using "E" to represent the electric field, which can be expressed numerically as the derivitave with respect to distance of the voltage, but thatoneguy is using "E" as defined as electromotive force, which is another name for voltage itself.

Logearav, thatoneguy is right in that the two "d"s in your equations are not algebraic symbols, but rather mean that the calculus operation of differentiation is being performed on the v and the x.

A way to say the equation verbally would be to say that the strength of the electric field can be expressed as the difference in voltage observed at two points very close together, divided by that distance, in units of volts per meter.

4. ### logearav Thread Starter Member

Aug 19, 2011
248
0
@davebee, Yes, E represents Electric field and d or x represents distance.
dV = -Edx so V = -Ex, if we integrate
So E = -V/x.
But how E = V/x? This formula is used for deriving Electric field of a parallel plate capacitor filled with dielectric medium.

Mar 6, 2009
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6. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
How, exactly did you integrate the equation? (and in relation to which variable?):

$\partial V =-\partial E\partial x$

You can't just cancel out the $\partial$ and end up with $\frac{V}{x}$

$\partial V$ represents a "slice" of V, but what is that "slice" related to? Current? Resistance? Distance? Field? Time?

$\partial E$ represents a "slice" of E, but in respect to what? Current? Resistance? Distance? Field? Time?

Maybe I need to go back to calc class.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Presumably logearav's question relates to the specific matter of the [uniform] electric field in a charged parallel plate capacitor. Given the field is linear / uniform for this particular case the differential form of the relationship between E,d & V may be recast in the simple non-differential from, since dV/dx is the same anywhere in the field. The variable x being considered to be displacement along the orthogonal axis between the plates.

The second (prime??) issue for the OP seems to be reconciling the sign convention adopted for the field "direction". The previous link I gave to Wikipedia 'explains' this in relation to the force experienced by a positive charge under the influence of the electric field.

8. ### logearav Thread Starter Member

Aug 19, 2011
248
0
@t_n_k, Thanks for the reply. This wikipedia article says E = -V/x.
Please see my attachment which depicts the derivation of capacitance of a parallel plate capacitor filled with dielectric medium. This derivation uses the formula V = Ex, while calculating the potential , when work is being done in moving charge in space filled with air and in space filled with dielectric medium. Thats why i raised the query.

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9. ### logearav Thread Starter Member

Aug 19, 2011
248
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Please see my attachment scanned from my text book which say dV = - Edx,
so ∫dV = -E∫dx which gives V = -Ex.
But while deriving the capacitance of a parallel plate capacitor filled with dielectric medium, the formula V = Ex is being used, for which i also attached images in the reply to t_n_k.

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10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The minus sign simply alerts one to the situation that the electric field points in the direction of decreasing electric potential. The direction an unbounded (free) positive charge would move in the presence of the field.

I note the derivation in your attachment doesn't explicitly note that convention. It might be regarded as a small anomaly but the outcome of the derivation would be the same irrespective of the implied direction of the electric field. It would hardly make any sense for instance if the capacitance were deduced to be a negative value. Perhaps one simply has to correctly 'guess' in which direction the hypothetical unit positive test charge is 'moved' to acquire the stated positive increase in electric potential.

11. ### BillO Distinguished Member

Nov 24, 2008
985
136
$

I\ think\ you\ are\ the\ victim\ of\ short\ hand\ notation.\\
\
\
First\ we\ have:\\
\
\
E=\frac{V}{X}\\
\
\
Where:\\
E\ is\ the\ overall\ electric\ field\ between\ two\ points\\
V\ is\ the\ measured\ voltage\ between\ those\ two\ points\\
and,\ X\ is\ the\ distance\ between\ those\ two\ points\\
\
\
This\ is\ only\ really\ useful\ with\ a\ uniform\ field\ like\ between\ the\\
plates\ of\ a\ capacitor.\\
\
\
Then\ we\ have:\\
\
\
E=-\frac{dv}{dx}\\
\
\
Where:\\
E\ is\ the\ electric\ field\ strength\ at\ a\ point\ (let's\ say\ x_{0})\\
v\ is\ the\ voltage\ as\ a\ function\ of\ x\\
and,\ dx\ is\ the\ differential\ infinitesimal\ around\ x_{0}\\
\
\
And\ should\ probably\ be\ more\ correctly\ written\ in\ general\ form\ as:
\
\
E(x)=-\frac{dv(x)}{dx}\\
\
\
So,\ given\ all\ that,\ is:\ \frac{V}{X}=-\frac{dv(x)}{dx}\ ?\\
\
\
Yes,\ when\ E\ is\ a\ uniform\ field\ like\ between\ the\\
plates\ of\ a\ capacitor\ such\ that\ \frac{dv(x)}{dx}\ is\ a\ constant\\

$

Last edited: Dec 13, 2011
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12. ### logearav Thread Starter Member

Aug 19, 2011
248
0
BillO, what is the difference between overall electric field between two points and electric field strength at a point?

13. ### BillO Distinguished Member

Nov 24, 2008
985
136
I used overall electric field to describe the field between the plates of a capacitor. I use it to mean a constant field. An example would be:

$
E\ =\ 100\frac{V}{mm}
$

I used field strength at a point to describe a field that varies with x and could take on any variable form. An example would be:

$
E(x)\ =\ A+\frac{B}{x^{2}}

$

Where A and B are constants and:

$such\ that\ at\ a\ point\ x_0;
\
\
E_{_{x_0}}\ =\ A+\frac{B}{(x_0)^{2}}

$