5.3 Coulomb's law (Page 3/11)

University physics volume 2 Page 3 / 11

\vec{F} (r) = \frac{1}{4 π ε_{0}} Q \sum_{i = 1}^{N} \frac{q_{i}}{r_{i}^{2}} {\hat{r}}_{i} .

In this expression, Q represents the charge of the particle that is experiencing the electric force $\vec{F}$ , and is located at $\vec{r}$ from the origin; the $q_{i} ’s$ are the N source charges, and the vectors ${\vec{r}}_{i} = r_{i} {\hat{r}}_{i}$ are the displacements from the position of the i th charge to the position of Q . Each of the N unit vectors points directly from its associated source charge toward the test charge. All of this is depicted in [link] . Please note that there is no physical difference between Q and $q_{i}$ ; the difference in labels is merely to allow clear discussion, with Q being the charge we are determining the force on.

Eight source charges are shown as small spheres distributed within an x y z coordinate system. The sources are labeled q sub 1, q sub 2, and so on. Sources 1, 2, 4, 7 and 8 are shaded red and sources 3, 5, and 6 are shaded blue. A test charge is also shown, shaded in green and labeled as plus Q. The r vectors from each source to the test charge Q are shown as arrows with tails at the sources and heads at the test charge. The vector from q sub 1 to the test charge is labeled as r sub 1. The vector from q sub 2 to the test charge is labeled as r sub 2, and so on for all eight vectors. — The eight source charges each apply a force on the single test charge Q . Each force can be calculated independently of the other seven forces. This is the essence of the superposition principle.

(Note that the force vector ${\vec{F}}_{i}$ does not necessarily point in the same direction as the unit vector ${\hat{r}}_{i}$ ; it may point in the opposite direction, $− {\hat{r}}_{i}$ . The signs of the source charge and test charge determine the direction of the force on the test charge.)

There is a complication, however. Just as the source charges each exert a force on the test charge, so too (by Newton’s third law) does the test charge exert an equal and opposite force on each of the source charges. As a consequence, each source charge would change position. However, by [link] , the force on the test charge is a function of position; thus, as the positions of the source charges change, the net force on the test charge necessarily changes, which changes the force, which again changes the positions. Thus, the entire mathematical analysis quickly becomes intractable. Later, we will learn techniques for handling this situation, but for now, we make the simplifying assumption that the source charges are fixed in place somehow, so that their positions are constant in time. (The test charge is allowed to move.) With this restriction in place, the analysis of charges is known as electrostatics , where “statics” refers to the constant (that is, static) positions of the source charges and the force is referred to as an electrostatic force .

The net force from two source charges

Three different, small charged objects are placed as shown in [link] . The charges

q_{1}

and

q_{3}

are fixed in place;

q_{2}

is free to move. Given

q_{1} = 2 e

q_{2} = −3 e

, and

q_{3} = −5 e

, and that

d = 2.0 \times 10^{−7} m

, what is the net force on the middle charge

q_{2}

Three charges are shown in an x y coordinate system. Charge q sub 1 is at x=0, y=d. Charge q sub 2 is at x=2 d, y=0. Charge q sub 3 is at the origin. Force F 1 2 is exerted on charge q sub 2 and points up. Force F 2 3 is exerted on charge q sub 2 and points to the left. Force F is exerted on charge q sub 2 and points at an angle theta above the minus x direction. — Source charges $q_{1}$ and $q_{3}$ each apply a force on $q_{2}$ .

Strategy

We use Coulomb’s law again. The way the question is phrased indicates that

q_{2}

is our test charge, so that

q_{1}

and

q_{3}

are source charges. The principle of superposition says that the force on

q_{2}

from each of the other charges is unaffected by the presence of the other charge. Therefore, we write down the force on

q_{2}

from each and add them together as vectors.

Solution

We have two source charges

(q_{1}

and

q_{3}),

a test charge

(q_{2}),

distances

(r_{21}

and

r_{23}),

and we are asked to find a force. This calls for Coulomb’s law and superposition of forces. There are two forces:

\vec{F} = {\vec{F}}_{21} + {\vec{F}}_{23} = \frac{1}{4 π ε_{0}} [\frac{q_{2} q_{1}}{r_{21}^{2}} \hat{j} + (- \frac{q_{2} q_{3}}{r_{23}^{2}} \hat{i})] .

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Source: OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3

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