## Problems & Exercises

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

(b) one charge of $+q$

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

(a) $\overrightarrow{E}=4.36\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N/C},\phantom{\rule{0.25em}{0ex}}\mathrm{35.0\xba}$, below the horizontal.

(b) No

(a) 0.263 N

(b) If the charges are distributed over some area, there will be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force.

$\begin{array}{lll}F& =& k\frac{\left|{q}_{1}{q}_{2}\right|}{{r}^{2}}=\mathrm{ma}\Rightarrow a=\frac{k{q}^{2}}{m{r}^{2}}\\ & =& \frac{(9.00\times {10}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}/{\text{C}}^{2}){(1.60\times {10}^{\mathrm{\u201319}}\phantom{\rule{0.25em}{0ex}}\text{m})}^{2}}{(1.67\times {10}^{\mathrm{\u201327}}\phantom{\rule{0.25em}{0ex}}\text{kg}){(2.00\times {10}^{\mathrm{\u20139}}\phantom{\rule{0.25em}{0ex}}\text{m})}^{2}}\\ & =& 3.45\times {10}^{16}\phantom{\rule{0.25em}{0ex}}\text{m/}{\text{s}}^{2}\end{array}$

(a) 3.2

(b) If the distance increases by 3.2, then the force will decrease by a factor of 10 ; if the distance decreases by 3.2, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10.

(a) $1\text{.}\text{04}\times {\text{10}}^{-9}$ C

(b) This charge is approximately 1 nC, which is consistent with the magnitude of charge typical for static electricity

- 0.859 m beyond negative charge on line connecting two charges
- 0.109 m from lesser charge on line connecting two charges

(a) $6\text{.}\text{94}\times {\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}\text{C}$

(b) $6\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}\text{N/C}$

(a) $\text{300}\phantom{\rule{0.25em}{0ex}}\text{N/C}\phantom{\rule{0.25em}{0ex}}(\text{east})$

(b) $4\text{.}\text{80}\times {\text{10}}^{-\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}\phantom{\rule{0.25em}{0ex}}(\text{east})$

(a) $5\text{.}\text{58}\times {\text{10}}^{-\text{11}}\phantom{\rule{0.25em}{0ex}}\text{N/C}$

(b)the coulomb force is extraordinarily stronger than gravity

(a) $-6\text{.}\text{76}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{C}$

(b) $2\text{.}\text{63}\times {\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\phantom{\rule{0.25em}{0ex}}(\text{upward})$

(c) $2\text{.}\text{45}\times {\text{10}}^{-\text{18}}\phantom{\rule{0.25em}{0ex}}\text{kg}$

(a) The forces are balanced in the *x*-direction, so the net force is vertical. It is composed of the sum of the vertical components of the Coulomb force and the gravitational force. $\begin{array}{l}{F}_{y}=-2\left(\frac{k{q}_{1}{q}_{2}}{{r}^{2}}+G\frac{{m}_{1}{m}_{2}}{{r}^{2}}\right)\mathrm{cos}45\xb0=\\ -2\left(\frac{8.99\times {10}^{9}(1.00\times {10}^{-9})(2.00\times {10}^{-9})}{8}+6.67\times {10}^{-11}\frac{(10.0)(10.0)}{8}\right)\\ \mathrm{cos}45\xb0=-4.36\times {10}^{-9}\text{N}\end{array}$

(b) No, it is in a metastable position. Since it cannot move horizontally, it cannot traverse any part of the track.

(c) $\begin{array}{l}{F}_{x}=-\left(\frac{k{q}_{1}{q}_{2}}{4}+G\frac{{m}_{1}{m}_{2}}{4}\right)\frac{1}{2}+\left(\frac{k{q}_{1}{q}_{2}}{12}+G\frac{{m}_{1}{m}_{2}}{12}\right)\frac{\sqrt{3}}{2}=\\ -\left(\frac{8.99\times {10}^{9}(-1.00\times {10}^{-9})(2.00\times {10}^{-9})}{4}\frac{1}{2}-6.67\times {10}^{-11}\frac{(10.0)(10.0)}{4}\frac{1}{2}\right)\\ +\left(\frac{8.99\times {10}^{9}(-1.00\times {10}^{-9})(2.00\times {10}^{-9})}{12}\frac{\sqrt{3}}{2}-6.67\times {10}^{-11}\frac{(10.0)(10.0)}{12}\frac{\sqrt{3}}{2}\right)\text{N}=\text{2}\text{.60}\times {\text{10}}^{-9}\text{N}\end{array}$ $\begin{array}{l}{F}_{y}=-\left(\frac{k{q}_{1}{q}_{2}}{4}+G\frac{{m}_{1}{m}_{2}}{4}\right)\frac{\sqrt{3}}{2}-\left(\frac{k{q}_{1}{q}_{2}}{12}+G\frac{{m}_{1}{m}_{2}}{12}\right)\frac{1}{2}=-2.08\times {10}^{-9}\text{N}\\ -\left(\frac{8.99\times {10}^{9}(1.00\times {10}^{-9})(2.00\times {10}^{-9})}{4}\frac{\sqrt{3}}{2}-6.67\times {10}^{-11}\frac{(10.0)(10.0)}{4}\frac{\sqrt{3}}{2}\right)\\ -\left(\frac{8.99\times {10}^{9}(1.00\times {10}^{-9})(2.00\times {10}^{-9})}{12}\frac{1}{2}+6.67\times {10}^{-11}\frac{(10.0)(10.0)}{12}\frac{1}{2}\right)\text{N}=-6.36\times {\text{10}}^{-9}\text{N}\end{array}$

(d) Yes.

(e) It will reach $(1,-\sqrt{3)}$ before it will change direction.

(f) The metastable positions where there is no component in one direction are $(0,0)$, $(4,0)$, $(2,2)$, and $(2,-2)$. They number 4.

## Test Prep for AP® Courses

(a) -0.1 C, (b) 1.1 C, (c) Both charges will be equal to 1 C, law of conservation of charge, (d) 0.9 C

a) Ball 1 will have positive charge and Ball 2 will have negative charge. b) The negatively charged rod attracts positive charge of Ball 1. The electrons of Ball 1 are transferred to Ball 2, making it negatively charged. c) If Ball 2 is grounded while the rod is still there, it will lose its negative charge to the ground. d) Yes, Ball 1 will be positively charged and Ball 2 will be negatively charge.

(a) 3.60×10^{10} N, (b) It will become 1/4 of the original value; hence it will be equal to 8.99×10^{9} N

(a) i) Field vectors near objects point toward negatively charged objects and away from positively charged objects.

(a) ii) The vectors closest to *R* and *T* are about the same length and start at about the same distance. We have that ${q}_{R}/{d}^{2}={q}_{T}/{d}^{2}$, so the charge on *R* is about the same as the charge on *T*. The closest vectors around *S* are about the same length as those around *R* and *T*. The vectors near *S* start at about 6 units away, while vectors near *R* and *T* start at about 4 units. We have that ${q}_{R}/{d}^{2}={q}_{S}/{D}^{2}$, so ${q}_{S}/{q}_{R}={D}^{2}/{d}^{2}=36/16=2.25$, and so the charge on *S* is about twice that on *R* and *T*.

(b)

(c)

$E=k\left[-\frac{q}{{(d+x)}^{2}}+\frac{2q}{{(x)}^{2}}+\frac{q}{{(d-x)}^{2}}\right]$

(d) The statement is not true. The vector diagram shows field vectors in this region with nonzero length, and the vectors not shown have even greater lengths. The equation in part (c) shows that, when $0<x<d$, the denominator of the negative term is always greater than the denominator of the third term, but the numerator is the same. So the negative term always has a smaller magnitude than the third term and since the second term is positive the sum of the terms is always positive.