# 18.7 Conductors and electric fields in static equilibrium  (Page 5/11)

 Page 5 / 11

(a) Using the symmetry of the arrangement, show that the electric field at the center of the square in [link] is zero if the charges on the four corners are exactly equal. (b) Show that this is also true for any combination of charges in which ${q}_{a}={q}_{d}$ and ${q}_{b}={q}_{c}$

(a) What is the direction of the total Coulomb force on $q$ in [link] if $q$ is negative, ${q}_{a}={q}_{c}$ and both are negative, and ${q}_{b}={q}_{c}$ and both are positive? (b) What is the direction of the electric field at the center of the square in this situation?

Considering [link] , suppose that ${q}_{a}={q}_{d}$ and ${q}_{b}={q}_{c}$ . First show that $q$ is in static equilibrium. (You may neglect the gravitational force.) Then discuss whether the equilibrium is stable or unstable, noting that this may depend on the signs of the charges and the direction of displacement of $q$ from the center of the square.

If ${q}_{a}=0$ in [link] , under what conditions will there be no net Coulomb force on $q$ ?

In regions of low humidity, one develops a special “grip” when opening car doors, or touching metal door knobs. This involves placing as much of the hand on the device as possible, not just the ends of one’s fingers. Discuss the induced charge and explain why this is done.

Tollbooth stations on roadways and bridges usually have a piece of wire stuck in the pavement before them that will touch a car as it approaches. Why is this done?

Suppose a woman carries an excess charge. To maintain her charged status can she be standing on ground wearing just any pair of shoes? How would you discharge her? What are the consequences if she simply walks away?

## Problems&Exercises

Sketch the electric field lines in the vicinity of the conductor in [link] given the field was originally uniform and parallel to the object’s long axis. Is the resulting field small near the long side of the object?

Sketch the electric field lines in the vicinity of the conductor in [link] given the field was originally uniform and parallel to the object’s long axis. Is the resulting field small near the long side of the object?

Sketch the electric field between the two conducting plates shown in [link] , given the top plate is positive and an equal amount of negative charge is on the bottom plate. Be certain to indicate the distribution of charge on the plates.

Sketch the electric field lines in the vicinity of the charged insulator in [link] noting its nonuniform charge distribution. A charged insulating rod such as might be used in a classroom demonstration.

What is the force on the charge located at $x=8.00 cm$ in [link] (a) given that $q=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ ? (a) Point charges located at 3.00, 8.00, and 11.0 cm along the x -axis. (b) Point charges located at 1.00, 5.00, 8.00, and 14.0 cm along the x -axis.

(a) Find the total electric field at $x=1.00 cm$ in [link] (b) given that $q=5.00 nC$ . (b) Find the total electric field at $x=11.00 cm$ in [link] (b). (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a single charge, double charge, etc., and what will its value(s) be?)

(a) ${E}_{x=\text{1.00 cm}}=-\infty$

(b) $2\text{.}\text{12}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}$

(c) one charge of $+q$

(a) Find the electric field at $x=5.00 cm$ in [link] (a), given that $q=1.00\phantom{\rule{0.25em}{0ex}}\text{μC}$ . (b) At what position between 3.00 and 8.00 cm is the total electric field the same as that for $–2q$ alone? (c) Can the electric field be zero anywhere between 0.00 and 8.00 cm? (d) At very large positive or negative values of x , the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of 11.0 cm is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight in this problem.)

(a) Find the total Coulomb force on a charge of 2.00 nC located at $x=4.00 cm$ in [link] (b), given that $q=1.00\phantom{\rule{0.25em}{0ex}}\text{μC}$ . (b) Find the x -position at which the electric field is zero in [link] (b).

(a) 0.252 N to the left

(b) $x=6.07 cm$

Using the symmetry of the arrangement, determine the direction of the force on $q$ in the figure below, given that ${q}_{a}={q}_{b}\text{=+}7\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{μC}$ and ${q}_{c}={q}_{d}=-7\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{μC}$ . (b) Calculate the magnitude of the force on the charge $q$ , given that the square is 10.0 cm on a side and $q=2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ .

(a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in [link] , given that ${q}_{a}={q}_{b}=-1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ and ${q}_{c}={q}_{d}\text{=+}1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ . (b) Calculate the magnitude of the electric field at the location of $q$ , given that the square is 5.00 cm on a side.

(a)The electric field at the center of the square will be straight up, since ${q}_{a}$ and ${q}_{b}$ are positive and ${q}_{c}$ and ${q}_{d}$ are negative and all have the same magnitude.

(b) $2\text{.}\text{04}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{N/C}\phantom{\rule{0.25em}{0ex}}\left(\text{upward}\right)$

Find the electric field at the location of ${q}_{a}$ in [link] given that ${q}_{b}={q}_{c}={q}_{d}\text{=+}2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{nC}$ , $q=-1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{nC}$ , and the square is 20.0 cm on a side.

Find the total Coulomb force on the charge $q$ in [link] , given that $q=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ , ${q}_{a}=2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ , ${q}_{b}=-3\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ , ${q}_{c}=-4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ , and ${q}_{d}\text{=+}1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μC}$ . The square is 50.0 cm on a side.

$0\text{.}\text{102}\phantom{\rule{0.25em}{0ex}}\text{N},$ in the $-y$    direction

(a) Find the electric field at the location of ${q}_{a}$ in [link] , given that ${q}_{\text{b}}=+10.00\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ and ${q}_{\text{c}}=–5.00\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ . (b) What is the force on ${q}_{a}$ , given that ${q}_{\text{a}}=+1.50\phantom{\rule{0.25em}{0ex}}\text{nC}$ ? Point charges located at the corners of an equilateral triangle 25.0 cm on a side.

(a) Find the electric field at the center of the triangular configuration of charges in [link] , given that ${q}_{a}\text{=+}2\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{nC}$ , ${q}_{b}=-8\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{nC}$ , and ${q}_{c}\text{=+}1\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{nC}$ . (b) Is there any combination of charges, other than ${q}_{a}={q}_{b}={q}_{c}$ , that will produce a zero strength electric field at the center of the triangular configuration?

(a) $\stackrel{\to }{E}=4.36×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N/C},\phantom{\rule{0.25em}{0ex}}35.0º$ , below the horizontal.

(b) No

#### Questions & Answers

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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It is the opposite of kinetic friction
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static fiction is friction between two surfaces in contact an none of sliding over on another, while Kinetic friction is friction between sliding surfaces in contact.
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I don't get it,if it's static then there will be no friction.
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Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration. To change from weight to mass divide by gravity (9.8 m/s^2).
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