1. **Problem:** Two charges $q_1 = -0.25 \times 10^{-6} C$ at $x=0$ and $q_2 = 6 \times 10^{-6} C$ at $x=1\,m$ on x-axis. Find point $x$ where electric field is zero. 2. **Solution:** Let point $x$ be distance from $q_1$ where electric field is zero. Electric field due to $q_1$ at $x$ is $E_1 = k|q_1|/x^2$, direction depends on charge. Electric field due to $q_2$ at $x$ is $E_2 = k|q_2|/ (1 - x)^2$ assuming $x < 1$. Since $q_1$ is negative, its field points towards it, $q_2$ positive field points away from it. For net field zero, $$ E_1 = E_2 $$ $$ \frac{|q_1|}{x^2} = \frac{|q_2|}{(1-x)^2} $$ $$ \sqrt{\frac{|q_1|}{|q_2|}} = \frac{x}{1-x} $$ Substitute values: $$ \sqrt{\frac{0.25 \times 10^{-6}}{6 \times 10^{-6}}} = \frac{x}{1-x} $$ $$ \sqrt{\frac{0.25}{6}} = \frac{x}{1-x} $$ $$ \frac{0.2041}{1} = \frac{x}{1-x} $$ Solving: $$ x = 0.2041(1 - x) $$ $$ x = 0.2041 - 0.2041x $$ $$ x + 0.2041x = 0.2041 $$ $$ x(1 + 0.2041) = 0.2041 $$ $$ x = \frac{0.2041}{1.2041} \approx 0.1694 m $$ **Answer:** The electric field is zero at approximately $0.169\,m$ to the right of $q_1$. --- 2(a). **Problem:** Electron velocity $v_0=40,000~m/s$, electric field $E=50~N/C$, time $t=1.5~ns = 1.5 \times 10^{-9}~s$. Find velocity $v$ after $t$. **Solution:** Electron charge $e=1.6 \times 10^{-19} C$, mass $m=9.11 \times 10^{-31} kg$. Acceleration: $$ a = \frac{eE}{m} = \frac{1.6 \times 10^{-19} \times 50}{9.11 \times 10^{-31}} = 8.78 \times 10^{12}~m/s^2 $$ Final velocity: $$ v = v_0 + at = 4.0 \times 10^{4} + 8.78 \times 10^{12} \times 1.5 \times 10^{-9} = 4.0 \times 10^{4} + 1.317 \times 10^{4} = 5.317 \times 10^{4}~m/s $$ 2(b). **Distance traveled:** $$ x = v_0 t + \frac{1}{2} a t^2 = 4.0 \times 10^{4} \times 1.5 \times 10^{-9} + \frac{1}{2} \times 8.78 \times 10^{12} \times (1.5 \times 10^{-9})^2 $$ $$ = 6.0 \times 10^{-5} + 9.87 \times 10^{-6} = 6.98 \times 10^{-5} m = 69.8 \mu m $$ --- 3(a). **Problem:** Electron released from rest, moves $4.5~m$ in $3~\mu s = 3 \times 10^{-6}~s$. Find magnitude and direction of $E$. **Solution:** Use $x = \frac{1}{2} a t^2$. $$ a = \frac{2x}{t^2} = \frac{2 \times 4.5}{(3 \times 10^{-6})^2} = 1.0 \times 10^{12} m/s^2 $$ Acceleration related to electric field: $$ E = \frac{ma}{e} = \frac{9.11 \times 10^{-31} \times 1.0 \times 10^{12}}{1.6 \times 10^{-19}} = 5.69 \times 10^{0} = 5.69 N/C $$ Direction: Since electron moves upward, $E$ points upward (electron force opposite to $E$ so $E$ is upward). 3(b). **Gravity effect check:** Gravitational acceleration $g = 9.8 m/s^2$ much less than electric acceleration $a = 10^{12} m/s^2$. Ratio: $$ \frac{a}{g} = \frac{10^{12}}{9.8} \approx 10^{11} $$ We can safely ignore gravity. --- 4(a). **Problem:** Particle mass $m=1.45~g=1.45 \times10^{-3} kg$, electric field $E = 650~N/C$ downward. Find charge $q$ to remain stationary. **Solution:** For stationary: $$ mg = qE $$ $$ q = \frac{mg}{E} = \frac{1.45 \times 10^{-3} \times 9.8}{650} = 2.18 \times 10^{-5} C $$ Charge sign must be positive to balance downward electric force by upward weight. 4(b). **Proton in electric field:** Weight $mg = eE$. $$ E = \frac{mg}{e} = \frac{1.67 \times 10^{-27} \times 9.8}{1.6 \times 10^{-19}} = 1.02 \times 10^{-7} N/C $$ --- 5. **Work to turn dipole 180°:** $E=46$, $p=3.02 \times 10^{-25}$ C.m, initial angle $\theta_i=64^\circ$. Work done: $$ W = pE (\cos \theta_i - \cos \theta_f) = 3.02 \times 10^{-25} \times 46 \times (\cos 64^\circ - \cos 244^\circ) $$ $$ \cos 64^\circ = 0.4384 ; \cos 244^\circ = -0.4384 $$ $$ W = 3.02 \times 10^{-25} \times 46 \times (0.4384 + 0.4384) = 3.02 \times 10^{-25} \times 46 \times 0.8768 = 1.22 \times 10^{-23} J $$ --- 6(a). **Ammonia dipole $p=5 \times 10^{-30} C.m$, $E=1.6 \times 10^6$. Change orientation parallel to perpendicular:** Change in potential energy: $$ \Delta U = pE (\cos 90^\circ - \cos 0^\circ) = pE (0 - 1) = -pE = -5 \times 10^{-30} \times 1.6 \times 10^6 = -8 \times 10^{-24} J $$ Absolute value is $8 \times 10^{-24} J$. 6(b). **Temperature at which average translational kinetic energy equals $\Delta U$:** $$ \frac{3}{2} k T = 8 \times 10^{-24} $$ Boltzmann constant $k = 1.38 \times 10^{-23}$ $$ T = \frac{2 \times 8 \times 10^{-24}}{3 \times 1.38 \times 10^{-23}} = 0.386 K $$ --- 7. **Charged rod length 0.14m, charge $-22 \mu C$, find E at 0.36m along axis.** Use formula for field on axis of rod: $$ E = \frac{1}{4\pi \epsilon_0} \frac{Q}{a b} \left( \frac{b}{\sqrt{b^2 + L^2/4}} - \frac{a}{\sqrt{a^2 + L^2/4}} \right) $$ Here, $a=0.36 - 0.07=0.29$, $b=0.36 +0.07=0.43$ m, $L=0.14$, $Q=-22 \times 10^{-6}$ C. Calculate numeric value: $$ k = 8.99 \times 10^9 $$ $$ E = 8.99 \times 10^9 \times \frac{22 \times 10^{-6}}{0.43 - 0.29} \left( \frac{0.43}{\sqrt{0.43^2 + 0.07^2}} - \frac{0.29}{\sqrt{0.29^2 + 0.07^2}} \right) $$ Compute: Numerator denominator factor approx $0.14$ Evaluate inside parentheses: $$ \frac{0.43}{\sqrt{0.1849 + 0.0049}}=\frac{0.43}{0.4335}=0.992 $$ $$ \frac{0.29}{\sqrt{0.0841 + 0.0049}}=\frac{0.29}{0.293} = 0.99 $$ Difference $=0.002$ So: $$ E = 8.99 \times 10^9 \times 22 \times 10^{-6} / 0.14 \times 0.002 = 8.99 \times 10^9 \times 1.5714 \times 10^{-4} \times 0.002 = 2.83 \times 10^{3} N/C $$ Direction: same as charge sign is negative, field points towards rod. --- 8. **Find $z$ where field along central axis of charged disk is half value at center.** Field at center: $$ E_0 = 2 \pi k \sigma $$ Field at distance $z$: $$ E_z = 2 \pi k \sigma \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right) $$ Set: $$ E_z = \frac{1}{2} E_0 \Rightarrow 1 - \frac{z}{\sqrt{z^2 + R^2}} = \frac{1}{2} $$ $$ \frac{z}{\sqrt{z^2 + R^2}} = \frac{1}{2} $$ Square both sides: $$ \frac{z^2}{z^2 + R^2} = \frac{1}{4} $$ $$ 4 z^2 = z^2 + R^2 $$ $$ 3 z^2 = R^2 \Rightarrow z = \frac{R}{\sqrt{3}} = \frac{0.6}{1.732} = 0.346 m $$ --- 9. **Infinite wire with linear charge density $\lambda=1.5 \times 10^{-10} C/m$. Find distance where $E=2.5 N/C$.** Field from wire: $$ E = \frac{2k\lambda}{r} $$ Solve for $r$: $$ r = \frac{2k \lambda}{E} = \frac{2 \times 8.99 \times 10^9 \times 1.5 \times 10^{-10}}{2.5} = 10.8 m $$ --- 10. **Three charges $+q$ at $30^\circ$, $+q$ at $150^\circ$, $-2q$ at $270^\circ$ on circle radius $r$. Find resultant field at center symbolically.** Field magnitude from point charge at center: $$ E = \frac{k q}{r^2} $$ Vector sum: $$ \vec{E} = \vec{E}_{30} + \vec{E}_{150} + \vec{E}_{270} $$ Components: $$ E_x = E (\cos 30^\circ + \cos 150^\circ + 0) = E (\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + 0) = 0 $$ $$ E_y = E (\sin 30^\circ + \sin 150^\circ - 2) = E (\frac{1}{2} + \frac{1}{2} - 2) = E (-1) $$ Result: $$ \vec{E} = - E \hat{y} = - \frac{kq}{r^2} \hat{y} $$ This means the resultant field points downward with magnitude $\frac{kq}{r^2}$. --- Final Answers: 1. $x \approx 0.169\, m$ 2a. $v = 5.3 \times 10^{4} m/s$ 2b. $x = 69.8 \mu m$ 3a. $E = 5.69~N/C$ upward 3b. Gravity negligible 4a. $q = +2.18 \times 10^{-5} C$ 4b. $E = 1.02 \times 10^{-7} N/C$ 5. $W = 1.22 \times 10^{-23} J$ 6a. $\Delta U = -8 \times 10^{-24} J$ 6b. $T = 0.386 K$ 7. $E \approx 2.83 \times 10^{3} N/C$ toward rod 8. $z = 0.346 m$ 9. $r = 10.8 m$ 10. $\vec{E} = - \frac{k q}{r^2} \hat{y}$ downward