Chapter 18

(a) $1.25\times {\text{10}}^{\text{10}}$

(b) $3.13\times {\text{10}}^{\text{12}}$

-600 C

$1.03\times {10}^{12}$

$9\text{.}\text{09}\times {\text{10}}^{-\text{13}}$

$1\text{.}\text{48}\times {\text{10}}^8\text{C}$

(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.

The separation decreased by a factor of 5.

$\begin{array}{lll}F& =& k\frac{\left|q_1{q}_2\right|}{r^2}=\mathrm{ma}\Rightarrow a=\frac{k{q}^2}{m{r}^2}\\ & =& \frac{(9.00\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2){(1.60\times {10}^{\mathrm{–19}}\text{m})}^2}{(1.67\times {10}^{\mathrm{–27}}\text{kg}){(2.00\times {10}^{\mathrm{–9}}\text{m})}^2}\\ & =& 3.45\times {10}^{16}\text{m/}{\text{s}}^2\end{array}$

(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

$1\text{.}\text{02}\times {\text{10}}^{-\text{11}}$

1. 0.859 m beyond negative charge on line connecting two charges
2. 0.109 m from lesser charge on line connecting two charges

$8\text{.}\text{75}\times {\text{10}}^{-4}$ N

(a) $6\text{.}\text{94}\times {\text{10}}^{-8}\text{C}$

(b) $6\text{.}\text{25}\text{N/C}$

(a) $\text{300}\text{N/C}(\text{east})$

(b) $4\text{.}\text{80}\times {\text{10}}^{-\text{17}}\text{N}(\text{east})$

(a) $2\text{.}\text{12}\times {\text{10}}^5\text{N/C}$

(b) one charge of $+q$

(a) 0.252 N to the left

(b) $x=\mathrm{6.07\; cm}$

(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) $2\text{.}\text{04}\times {\text{10}}^7\text{N/C}(\text{upward})$

$0\text{.}\text{102}\text{N},$ in the $-y$$$$$$$ direction

(a) $\overrightarrow{E}=4.36\times {\text{10}}^3\text{N/C},\mathrm{35.0º}$, below the horizontal.

(b) No

(a) $5\text{.}\text{58}\times {\text{10}}^{-\text{11}}\text{N/C}$

(b)the coulomb force is extraordinarily stronger than gravity

(a) $-6\text{.}\text{76}\times {\text{10}}^5\text{C}$

(b) $2\text{.}\text{63}\times {\text{10}}^{\text{13}}{\text{m/s}}^2(\text{upward})$

(c) $2\text{.}\text{45}\times {\text{10}}^{-\text{18}}\text{kg}$

The charge $q_2$ is 9 times greater than $q_1$.

(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. $\begin{array}{l}{F}_y=-2\left(\frac{k{q}_1{q}_2}{r^2}+G\frac{m_1{m}_2}{r^2}\right)\cos 45°=\\ -2\left(\frac{8.99\times {10}^9(1.00\times {10}^{-9})(2.00\times {10}^{-9})}{8}+6.67\times {10}^{-11}\frac{(10.0)(10.0)}{8}\right)\\ \cos 45°=-4.36\times {10}^{-9}\text{ N}\end{array}$

(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.