Paper 3 Electric Fields Answers

Q1 Charged Particle Deflection

1. $F=qE=(1.60\times 10^{-19})(3.0\times 10^4)=4.8\times 10^{-15}\mathrm{N}$, so $a=F/m=2.87\times 10^{12}\mathrm{m}{\mathrm{s}}^{-2}$ upward. [3]
2. $t=L/v=0.040/(2.4\times 10^6)=1.67\times 10^{-8}\mathrm{s}$. [2]
3. $y=\frac{1}{2}a{t}^2=0.5(2.87\times 10^{12})(1.67\times 10^{-8}{)}^2=4.0\times 10^{-4}\mathrm{m}$. [3]
4. Horizontal velocity is constant while vertical acceleration is constant, so the path is parabolic. [2]

Q2 Point-Charge Graph Reasoning

1. Potential is energy per unit charge and adds algebraically; field has direction as force per unit positive charge. [2]
2. $V$ decreases as $1/r$ while remaining positive; $E$ decreases faster as $1/r^2$ and is directed radially outward. [3]
3. $V$ becomes negative; $\vec{E}$ reverses direction, while field magnitude still follows $1/r^2$. [2]