4.4 Calculating electric field from potential from Office of Academic Technologies on Vimeo.

- Example 1: Calculating electric field of a disc charge from its potential
- Example 2: Calculating electric field of a ring charge from its potential

**4.4 Calculating electric field from potential**

Earlier we have studied how to find the potential from the electric field. Calculating potential from E field was directed from the definition of potential, which led us to an expression such that potential difference between two points is equal to minus integral of E dot dl, integrated from initial to that final point. So from this expression if we knew the electric field we could easily calculate the potential difference that charge will experience whenever it is moved along a specific path from an original point to a final point.

Now we will ask the opposite question and we will say, “Can we calculate the electric field from the potential?” The answer to that is, yes. Therefore our title is Calculating Electric Field from Potential.

In this case we assume we know the potential at every point in the region interest. Knowing potential in the region of interest means we know all the equipotential surfaces in that region. Let’s represent those equipotential surfaces from the cross-sectional point of view, something like this. Therefore these quantities or these represent, these lines represent the cross section of these equipotential surfaces.

Let’s assume that we move a charge from one equipotential surface to another one along a specific path. We also know the electric field lines are always perpendicular to the equipotential surfaces, therefore these angles are 90 degrees for these equipotential surfaces they have the voltage of let’s say v1, v2, v3, and so on and so forth.

Let’s assume that we move our charge from one equipotential surface towards the other one along this path. Let’s call that displacement vector as l and therefore dl will represent an incremental displacement vector along this path. Let’s denote the angle between E and dl as theta.

All right now, we have seen that the voltage was defined as, or the potential was defined as negative of the work done in moving the charge from infinity to the point of interest per unit test charge. From there work done is equal to minus q times, the charge times the potential.

Well, work done is also equal to F of dl–we can call this as the incremental work done is equal to F dot dl. Since Coulomb force acting a charge, let’s say a positive charge plus q, which is being moved from one equipotential surface to the other one is equal to q times E. The incremental work done by moving that charge when an incremental displacement of dl is going to be equal to q E dot dl. Or, in a more expressive form it will equal to, since it’s a dot product, q E magnitude dl magnitude, times cosine of the angle between these two vectors.

Well, using the definition of potential therefore we can say that the incremental work done is going to be equal to minus q times the incremental potential difference that it’s being moved through.

In this expression the left-hand side of these two expressions, the left-hand sides are equal to therefore we can easily equate the right-hand sides. If we do that we will have minus q dV is going to be equal to q E dl times cosine of theta. Since q is common on the left-hand side and the right-hand side we can divide both sides and eliminate the charge, and if we move dl to one side of the expression to collect the differential terms on one side then we will have E cosine of theta is equal to dV over dl, with a minus sign.

Right here, I’m going to write down this expression in a more general form in the form of partial [inaudible 08:23] equation rather than doing it in a fashion like this, because potential product can be a function of different coordinates and that’s going to give us then E cosine of theta is equal to minus delta V over delta S–I should say del V over del l.

Let’s try to integrate this E times cosine theta term. Well, if we extend the direction of this displacement vector and we take the projection of the electric field along this direction then we’re going to end up with the component of the electric field in the direction of this displacement vector. We can call that one as the electric field vector component in the direction of l.

Now let’s represent that as E sub l. So, this quantity over here gives us the component of electric field in the direction of displacement vector, vector l, so we can therefore state that negative rate of change of potential with distance in any direction gives the component of electric field in that direction.

This is a very important result and in a rectangular coordinate system, let’s say in the Cartesian coordinate system, therefore we can say that the x component of the electric field will be equal to minus partial derivative of potential function with this vector x coordinate, y component of the electric field will equal to partial derivative of potential with respect to y component, and finally the z component will be equal to partial derivative of potential with respect to z component.

Now we know that electric field is a vector quantity were as potential is a scalar quantity, so here through mathematical operation we are obtaining components of the vector from the scalar quantity. Instead of expressing all these three coordinates with separate equations we introduce a notation system through an operator, which is called del operator. In a rectangular coordinate system, in a Cartesian coordinate system it’s a partial differential operator and it’s equal to del over del x unit vector I, plus del over del y unit vector j, plus del over del z unit vector k.

So in terms of this notation we can express the electric field vector is equal to minus del operator acting on the potential function V. This operation is called Gradient of V or Gradient of Potential. So the negative gradient of potential gives us the electric field vector.

You might not be familiar with the partial differentiation so far. As a matter of fact, partial differentiation is not really different than the total differentiation. If a function is a function of different variables, if we’re the partial derivative in respect to a specific variable, we simply take the other variables as constant during that process.

If we do a quick example related to this: Let’s assume that we have a potential, which is a function of x, x, and z coordinates. So, let V is equal to x, y, z is equal to 2x squared y cubed z, minus 3y squared z, plus 6xy, z cubed.

Now let’s assume that the potential function in a given region is varying according to this mathematical function. We would like to figure out the corresponding electric field in that region. The x component of the electric field is negative partial derivative of this potential function with respect to x, so that’s going to be equal to minus. We’ll take the derivative of this function with respect to x and in doing that we’ll keep the y and z as constant, so the first one is going to give us 4x and we’re keeping y and z constant, so we’ll have y cubed z, plus we’ll take the derivative with respect to x.

The second term is going to give us zero because there is no x dependence in that term and we’re turning y and z constant therefore we’re going to end up with zero from here, and the next one will give us plus 6yz cubed. So this would be the x component of the electric field vector.

Similarly y component will be minus del V over del y, which is going to be equal minus–again now we will take the derivative with respect to y and we will keep x and z constant during the process–and derivative with respect to y will be 3 times 2 is 6x square y square z and then minus, we’ll have 6yz for the second term, and plus 6xz cubed for the last term, once we take the derivative with respect to y.

Finally the x component is going to be equal to minus del v over del z which is going to be equal to minus–now we’re going to take the derivative with respect z, keeping x and y constant–the first one will give us 2x square y cubed. The second term will give us minus 3y squared, and the last term is going to be plus 18xyz squared.

So now we know the components of the electric field vector. The electric field vector therefore is going to be equal to Exi plus Eyj plus Exk.

Of course the magnitude of the electric field vector will be equal to Ex squared, plus Ey squared, plus Ez squared, in a square root.

So once we know the potential function then we can easily calculate the corresponding electric field components simply by taking what we call the negative gradient of that potential function.

Now, if you’re interested with the value of the electric field at specific points or for specific x and y and z then we simply substitute those values for x, y, and z in order to get specific value of the electric field at those specific points.