Special Cases in Electrostatics - Fluids - MCAT Physics and Math Review

MCAT Physics and Math Review

Chapter 5: Electrostatics and Magnetism

5.5 Special Cases in Electrostatics

In this section, we will explore some of the unique setups in electrostatics that are common on the MCAT.


An equipotential line is a line on which the potential at every point is the same. That is, the potential difference between any two points on an equipotential line is zero. Drawn on paper, equipotential lines may look like concentric circles surrounding a source charge. In three-dimensional space, these equipotential lines would actually be spheres surrounding the source charge. From the equation for electrical potential, we can see that no work is done when moving a test charge q from one point on an equipotential line to another. Work will be done in moving a test charge qfrom one line to another, but the work depends only on the potential difference of the two lines and not on the pathway taken between them. This is entirely analogous to the displacement of an object horizontally on a level surface. Because the object’s height above the ground has not changed, its gravitational potential energy is unchanged. Furthermore, a change in the object’s gravitational potential energy will not depend on the pathway taken from one height to another but only on the actual vertical displacement.


Because the work to move a charge from one equipotential line to another does not depend on the path between them, we know that we are dealing only with conservative forces when moving the charge. Conservative and nonconservative forces are discussed in Chapter 1 of MCAT Physics and Math Review.


In the diagram below, an electron goes from point a to point b in the vicinity of a very large positive charge. The electron could be made to follow any of the paths shown. Which path requires the least work to get the electron charge from a to b?


As stated, the work depends only on the potential difference and not on the path, so any of the paths shown would require the same amount of work in moving the electron from a to b. Note that because the source charge is positive, point b is at a lower electrical potential than point a. However, because the test charge is negative, the electrical potential energy is higher at point b than point a. This should make sense: the electron will have to gain energy to be moved farther away from positive source charge.


Much of the reactivity of organic compounds is based on separation of charge. The electric dipole, which results from two equal and opposite charges being separated a small distance d from each other, can be transient (as in the case of the moment-by-moment changing distribution of electrons in the electron cloud of an atom or molecule that create London dispersion forces) or permanent (as in the case of the molecular dipole of water or the carbonyl functional group).

The electric dipole can be visualized as a barbell: the equal weights on either end of the bar represent the equal and opposite charges separated by a small distance, represented by the length of the bar. We’ll analyze the generic dipole in Figure 5.3 and then work through the specific example of one of the most important electric dipoles, the water molecule.

Figure 5.3. A Generic Dipole

The dipole in Figure 5.3 has charges +q and –q separated by a distance d. Notice that +q and –q are source charges, even though they are written in lowercase. Given the dipole, we may want to calculate the electrical potential at some point P near the dipole. The distance between the point in space and +q is r1; the distance between the point in space and –q is r2; the distance between the point in space and the midpoint of the dipole is r. We know that for a collection of charges, the electrical potential P is the scalar sum of the potentials due to each charge at that point. In other words,

For points in space relatively distant from the dipole (compared to d), the product of r1 and r2 is approximately equal to the square of r, and r1r2 is approximately equal to d cos θ. When we plug these approximations into the equation above, we get

Equation 5.7

The product of charge and separation distance is defined as the dipole moment (p) with SI units of C · m:

p = qd

Equation 5.8

The dipole moment is a vector, but its direction is defined differently by physicists and chemists. Physicists define the vector along the line connecting the charges (the dipole axis), with the vector pointing from the negative charge toward the positive charge. Chemists usually reverse this convention, having p point from the positive charge toward the negative charge. Sometimes, chemists draw a crosshatch at the tail end of the dipole vector to indicate that the tail end is at the positive charge.


The H2O molecule has a dipole moment of 1.85 D. Calculate the electrical potential due to a water molecule at a point 89 nm away along the axis of the dipole. (Note:

and 1 D (debye) = 3.34 × 10−30 C · m)


Because the question asks for the potential along the axis of the dipole, the angle θ is 0°. Substitute the values into the equation for the dipole potential and multiply 1.85 D by 3.34 × 10−30 to convert it to C · m:

One very important equipotential line to be aware of is the plane that lies halfway between +q and –q, called the perpendicular bisector of the dipole. Because the angle between this plane and the dipole axis is 90° (and cos 90° = 0), the electric potential at any point along this plane is 0. The magnitude of the electric field on the perpendicular bisector of the dipole can be approximated as

Equation 5.9

The electric field vectors at the points along the perpendicular bisector will point in the direction opposite to p (as defined directionally by physicists).

The dipole is a classic example of a setup upon which torques can act. In the absence of an electric field, the dipole axis can assume any random orientation. However, when the electric dipole is placed in a uniform external electric field, each of the equal and opposite charges of the dipole will experience a force exerted on it by the field. Because the charges are equal and opposite, the forces acting on the charges will also be equal in magnitude and opposite in direction, resulting in a situation of translational equilibrium. There will be, however, a net torque about the center of the dipole axis:


The dipole is a great example of how the MCAT can test kinematics and dynamics in an electrostatics setting. For a dipole at some angle in an external electric field, there will be translational equilibrium, but not rotational equilibrium. This is because the forces are opposite directions (left and right in Figure 5.4), but the torques are in the same direction (clockwise for both).

Thus, the net torque on a dipole can be calculated from the equation

τ = pE sin θ

Equation 5.10

where p is the magnitude of the dipole moment (p = qd), E is the magnitude of the uniform external electric field, and θ is the angle the dipole moment makes with the electric field. This torque will cause the dipole to reorient itself so that its dipole moment, p, aligns with the electric fieldE, as shown in Figure 5.4.

Figure 5.4. Torque on a Dipole from an External Electric Field


The electric dipole is most likely to be tested qualitatively or in the context of a passage or reaction on Test Day. It is unlikely that these mathematical relations will be presented or tested without background.

MCAT Concept Check 5.5:

Before you move on, assess your understanding of the material with these questions.

1. Define the following terms:

· Equipotential lines:

· Electric dipole:

2. What is the voltage between two points on an equipotential line? Will this voltage cause a charge to move along the line?

3. Why is the electrical potential at points along the perpendicular bisector of a dipole 0?

4. What is the behavior of an electric dipole when exposed to an external electric field?