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# Electrostatics Study Guide (page 3)

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Updated on Sep 27, 2011

## Electric Field and Potential

As we discovered previously, there is no need for contact in order to have electrical interaction, and this is not the only case we learned about so far. Gravitational force is carried out by the gravitational field, which changes the properties of the space around an object: The more massive the object, the larger the gravitational field and the greater the interaction with other objects found in proximity. Heating though radiation is another noncontact interaction we discussed. In the case of radiation, electromagnetic waves produced by a far-away source, such as the sun, can heat up an object, effectively increasing its temperature.

The space around an electrically charged object carries special properties, and we call this an electrical field. Any other charge in this region will interact with the electrical field, and we call this charge a testcharge. Hence, we have both a field-producing charge and a test-charge(s) that will be affected by the electrical field.

The electric field that propagates electrical interaction has vector properties because a positive charge will affect the charges around it in a different way than a negative charge will. The direction of the electric field and of the electric force is the same at any point around the charge.

The vectors starting or finishing on the charges are called electrical field lines, and they represent the direction of the electric field and electric force, as seen in Figure 13.7. The density of electrical field lines in a region represents the strength of the field in that area.

Depending on the position of the test-charge, there are many arrangements of the field lines encountered in applications. One is shown in the following example and figure.

### The Electric Field

The electric field at a location around a charge q is defined to be proportional to the force at the specified location and inversely proportional to the charge by the field.

#### Example 1

Consider positive and negative spherical charges in proximity. Draw the field lines.

#### Solution 1

The field lines are similar to the ones we have shown previously, but because the two particles interact, the field created by each will affect the other particle. The final arrangement looks like that in Figure 13.8.

According to the definition for electric field, the unit for the electric field will be the Newton/Coulomb (N/C). Because force was previously defined by Coulomb's law, if we consider the charge ql creating a field E and a test-charge q2 to be present in the field, then the field at point r with respect to charge ql is:

where F21 is the force that charge q2 feels due to the field created by charge q1. In the last expression, the field created by the charge q1 is seen to be in the same direction as the vector position r if the charge is positive and opposite to r if the charge is negative, relative to the field lines drawn at the beginning of this section.

If we are interested in the value of the field created by charge q1 then:

If the field is created by two or more electrical charges, the vector expression is helpful in finding the net electric field at one location as we will see in the following example.

#### Example 2

Consider two charges, as shown in Figure 13.9, creating a net electric field. Find the direction and the value of the net field at point A. The charges are q1 = + 1.6 · 10–19 C and q2 = –1.6 · 10–19C, and the distance to the point is r1 = 22 cm and r2 = 12 cm.

### Electric Potential Energy

The electric potential energy (EPE) per unit of charge defines the electric potential

#### Solution 2

First, convert the data into SI units. Then, find the field determined by each charge respectively at point A, and next find the net field.

r1 = 22cm = 0.22m

r2 = 12 cm = 0.12 m

q1 = +1.6 · 10–19 C

q2 = –1.6 · 10–19 C

Enet = ?

E1 =

The vector position has both a horizontal and a vertical component, and they can be represented as:

r1 = (r1x, r1y) = (r1 · cos α ;r1 · sin α)

where α is the angle the vector makes with the positive x-axis. In the case of the vector r1 the angle is 360° – 60° = 300°

r1 = (r1x,r1y) = (r1 · cos 300°; r1 · sin 300°)

r1 = (r1x,r1y) = (0.22 · cos 300°; 0.22 · sin 300°)

r1 = (r1x,r1y) = (0.11;–0.19) m

E1 =

E1 = 1.4 · 10–7 – (0.11;–0.19) N/C

E1 = (0.15;–0.27) · 10–7 N/C

A similar calculation can be made for the components of the second charge distance and electric field, and it can be said that:

r2 = (r2x,r2y) = (r2x,r2y) = r2 · cos 0°;r2 · sin 0°)

r2 = (r2x,r2y) = (r2x,r2y) = 0.12 · cos 0°;0.12 · sin 0°)

r2 = (r2x,r2y) = (0.12;0) m

E2 = (1;0) · 10–7N/C

The net electric field is the vector sum of the two fields:

Enet = E1 + E2 = [(0.15;–0.27) + (1;0)] · 10–7N/C

Enet = E1 + E2 = (1.15;–0.27) · 10–7N/C

And the direction of the field can be found from the Ex and Ey components of the field:

θ = –13°

Similar to the situation of a moving object in a gravitational field (where a potential energy is associated with the position of the object in the field), an electric potential energy can be defined for a charge placed in an electric field. As a greater mass needs more energy to be moved in a gravitational field, so does a larger charge need more energy to be moved through an electric field.

Electric potential, known as V, is "measured" in joule/coulombs and is called a volt (from Alessandro Volta, 1745–1827). Due to the unit for electric potential, this quantity is also called voltage. The voltage is seen previously to depend on energy, which is a scalar quantity; therefore, voltage is also a scalar. Voltage will still be positive or negative depending on the type of charge. The quotation marks in the first sentence of this paragraph are intended to emphasize the fact that voltage based on electric potential energy is a relative quantity. Potential energy, if you remember from mechanics, is proportional to the position of an object but with respect to an origin. A more measurable quantity is the difference in potential energy, hence the difference of voltage between two positions.

If a charge q is moving in an electric field E between positions A and B, then the field does work on the charge that is equal to the variation of the electric potential energy (analogous to the work done by gravity: W = – ΔPE).

In this case, the work to move the charge from point A to B is:

W = – ΔEPE = –(EPEBEPEA)

Dividing both sides by the charge q:

And considering the previous definition of voltage:

The work can be related to the voltage difference by:

#### Example 3

Consider a test-charge q = –1.6 pC move in an electric field E from point A to point B. The work done by the field to move the charge is 2.8 fJ. Find the potential difference between the two points.

#### Solution 3

First, convert the units and then proceed in formulating the solution.

W = 2.8 fJ = 2.8 · 10–15J

q = –1.6 · 10–12 C

ΔV = ?

ΔV = 1.7 · 10–3

Practice problems of this concept can be found at: Electrostatics Practice Questions

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