Edexcel A-Level Physics Electric and Magnetic Fields

Negative because the direction of field is direction of force on a positive charge

Or

Field downwards means negatively charged Earth and negative repels negative

Or

Negative because the force is in the opposite direction to the electric field

(1)

Use of F=E Q

- Use of W=m g

- F-W to determine resultant force

- Use of F=m a

- $a=2.2 \mathrm{~m} \mathrm{~s}^{-2}$

Example of calculation
$$\begin{aligned} & F=120 \mathrm{~V} \mathrm{~m}^{-1} \times 3.00 \times 10^{-7} \mathrm{C}=3.60 \times 10^{-5} \mathrm{~N} & W=3.00 \times 10^{-6} \mathrm{~kg} \times 9.81 \mathrm{Nkg}^{-1}=2.94 \times 10^{-5} \mathrm{~N} & \text { Resultant force }=3.60 \times 10^{-5} \mathrm{~N}-2.94 \times 10^{-5} \mathrm{~N}=6.57 \times 10^{-6} \mathrm{~N} & a=6.57 \times 10^{-6} \mathrm{~N} \div 3.00 \times 10^{-6} \mathrm{~kg} & =2.19 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$$

(5)

Use $E=Q/(4\pi\varepsilon_0r^2)$ or $E=kQ/r^2$, and $A=4\pi r^2$.

$120=(8.99\times10^9)Q/(6.4\times10^6)^2$, so $Q=5.47\times10^5\,\mathrm{C}$. The surface area is $5.15\times10^{14}\,\mathrm{m^2}$, giving $Q/A=1.1\times10^{-9}\,\mathrm{C\,m^{-2}}$.

Use Coulomb's law, $F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q^2}{r^2}=k\dfrac{Q^2}{r^2}$. Thus $1.16\times10^{-3}=(8.99\times10^9)Q^2/(0.47)^2$, giving $Q=1.69\times10^{-7}\,\mathrm{C}$, about $2\times10^{-7}\,\mathrm{C}$.

Use $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}=kQ/r$. With the balloon negatively charged, $V=(8.99\times10^9)(-1.7\times10^{-7})/0.30=-5.09\times10^3\,\mathrm{V}$, about $-5100\,\mathrm{V}$.