Chapter 18: Electric Charge and Electric FieldContent check — College Physics, Chapter 18: Electric Charge and Electric Field

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

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

(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.

(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

(a) 0.859 m beyond negative charge on line connecting two charges

(b) 0.109 m from lesser charge on line connecting two charges

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

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

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

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

(c) $$\begin{array}{l}{F}_x=-\frac{k{q}_1{q}_2}{4}+G\frac{m_1{m}_2}{4}\frac{1}{2}+\frac{k{q}_1{q}_2}{12}+G\frac{m_1{m}_2}{12}\frac{\sqrt{3}}{2}=\\ -\frac{8.99\times {10}^9(-1.00\times {10}^{-9})(2.00\times {10}^{-9})}{4}\frac{1}{2}-6.67\times {10}^{-11}\frac{(10.0)(10.0)}{4}\frac{1}{2}\\ +\frac{8.99\times {10}^9(-1.00\times {10}^{-9})(2.00\times {10}^{-9})}{12}\frac{\sqrt{3}}{2}-6.67\times {10}^{-11}\frac{(10.0)(10.0)}{12}\frac{\sqrt{3}}{2}\text{N}=\text{2}\text{.60}\times {\text{10}}^{-9}\text{ N}\end{array}$$ $$\begin{array}{l}{F}_y=-\frac{k{q}_1{q}_2}{4}+G\frac{m_1{m}_2}{4}\frac{\sqrt{3}}{2}-\frac{k{q}_1{q}_2}{12}+G\frac{m_1{m}_2}{12}\frac{1}{2}=-2.08\times {10}^{-9}\text{ N}\\ -\frac{8.99\times {10}^9(1.00\times {10}^{-9})(2.00\times {10}^{-9})}{4}\frac{\sqrt{3}}{2}-6.67\times {10}^{-11}\frac{(10.0)(10.0)}{4}\frac{\sqrt{3}}{2}\\ -\frac{8.99\times {10}^9(1.00\times {10}^{-9})(2.00\times {10}^{-9})}{12}\frac{1}{2}+6.67\times {10}^{-11}\frac{(10.0)(10.0)}{12}\frac{1}{2}\text{N}=-6.36\times {\text{10}}^{-9}\text{ N}\end{array}$$

(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.