Electric potential - doubt in sign convention

[logearav]

Electric potential --- doubt in sign convention

Homework Statement

Revered members,
I know dV = -E.dx so
E = -dV/dx
So E should be E = -V/x . But while deriving capacitance of a parallel plate capacitor when dielectric medium is introduced we use E = V/x.
But how E = V/x?

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[NascentOxygen]

I know dV = -E.dx so
E = -dV/dx
So E should be E = -V/x .

Where have you seen a negative sign like this? If E is positive, then voltage increases with x. So E and dV/dx have the same sign.

 

[vkash]

Homework Statement

Revered members,
I know dV = -E.dx so
E = -dV/dx
So E should be E = -V/x . But while deriving capacitance of a parallel plate capacitor when dielectric medium is introduced we use E = V/x.
But how E = V/x?

When it comes to study of capacitances we are concerned to calculate charges on the capacitor not the direction in which potential increase or decrease.
However if you want to calculate the increase, decrease in potential with the help of electric field then you can with proper signs.
writing E=-V/x may create problem in soecases. so you should always us E=-dV/dx.

 

[ehild]

Where have you seen a negative sign like this? If E is positive, then voltage increases with x. So E and dV/dx have the same sign.

There are some confusions when using the term potential and voltage.

The electric field $\vec{E}$ is negative gradient of the potential U: $\vec{E}$=- grad(U). E points in the direction of potential drop: from positive to negative potential.

In a homogeneous electric field, parallel to the x axis,
E=-dU/dx. The integral of E between two points A and B is the work W done by the field on a unit positive charge when it moves from A to B:

$$W(A\rightarrow B)=\int_{x_A}^{x_B}{E dx}=-\int_{x_A}^{x_B}{(dU /dx)dx}$$
$$\int_{x_A}^{x_B}{E dx}=E(x_B-x_A)$$
$$-\int_{x_A}^{x_B}{(dU /dx)dx}=-\int_{x_A}^{x_B}{dU}=-U(x_B)+U(x_A)$$
$$W(A\rightarrow B)=E(x_B-x_A)=-U(x_B)+U(x_A)$$
If X_B-x_A= D ,
$$W=E D = U_A-U_B$$

The work done by he field when a unit positive charge moves from A to B is equal to the negative of the potential difference between points A and B. The work of the electric field is positive if U_A >U_B.

"Voltage" V is used in practical texts in the meaning of potential drop when going from A to B: It is the negative potential difference V(AB)=-(U_B-U_A)=-Δ U.

The potential drops on a resistor in the direction of the current flow from A to B. Ohm's law states that the voltage V measured between points A and B is V=IR. This means V=-ΔU=U_A-U_B=IR. Across a capacitor, the potential of the positive plate U_A minus potential of the negative plate U_B is equal to ED: U_A-U_B=-ΔU=ED, that is V=ED.

ehild

 

[logearav]

Thanks for the detailed explanation, ehild. What is the position of the two points A and B in a electric field? For example, i draw a straight line XY. Where should i position the points A and B that satisfies your explanation?
X_________________________________________y

 

[ehild]

When you choose a direction it means you go from the initial position to the final one. ΔU means the potential at the final position - the potential at the initial position.

ehild

 

[logearav]

Thanks ehild. Now, if i choose a point A say 5 cm from X and B at 9 cm from X, which is the point with higher potential?

 

[ehild]

The more positive one.

ehild

 

[logearav]

Thanks again. How could we find which point is more positive?

 

[ehild]

Thanks again. How could we find which point is more positive?

First you get information from the problem.
It is given either the charge (plus or minus) or the potential or the electric field somewhere. A positively charged object has electric field lines originating from the charges and pointing away from it. Do not forget that the direction of electric field is +-----> - The electric field points in the direction of decreasing potential. Using the information given, you can find out which point is positive or negative with respect to an other point, applying Coulomb's Law, Ohm's Law or the relation between potential and electric field E=-grad (U).

It has no sense asking which points of a line are positive without giving the electric details.


I suggest to read this page.

http://www.physicsclassroom.com/class/circuits/u9l1c.cfm

ehild

 

[logearav]

ok. Now i am confronted with the magnitude of test charge and source charge. Should they be of same magnitude? I presume the test charge is unit positive charge. But i can't get an idea of magnitude of source charge, in the concept of electric potential

 

[ehild]

A test charge is a little (positive) charge which is used to measure the electric field. It is so small that does not change the field. The source charge makes the electric field. The electric field has potential.

ehild

 

[logearav]

Thanks a lot ehild. The link you have given is so wonderful. I understood the concept. Thanks again.