Paper 2 Electric Fields Answers

Q1 Coulomb Force and Field

1. $F=k|Q_1{Q}_2|/r^2=8.99\times 10^9(3.0\times 10^{-9})(8.0\times 10^{-9})/(0.120{)}^2=1.50\times 10^{-5}\mathrm{N}$. [3]
2. Opposite charges attract. [1]
3. $E=kQ/r^2=8.99\times 10^9(3.0\times 10^{-9})/(0.120{)}^2=1.87\times 10^3\mathrm{N}{\mathrm{C}}^{-1}$. [2]

Q2 Uniform Field Between Plates

1. $E=V/d=1500/(6.0\times 10^{-3})=2.50\times 10^5\mathrm{V}{\mathrm{m}}^{-1}$. [2]
2. $F=qE=(1.60\times 10^{-19})(2.50\times 10^5)=4.0\times 10^{-14}\mathrm{N}$. [2]
3. Opposite to the electric field because the electron is negative. [1]

Q3 Potential and Energy

1. $\Delta U=q\Delta V=(2.0\times 10^{-6})(35-120)=-1.70\times 10^{-4}\mathrm{J}$. [3]
2. Yes. The potential energy decreases, so the field does positive work of $1.70\times 10^{-4}\mathrm{J}$. [2]
3. It is energy per unit charge and has no direction; signs are handled algebraically. [1]