Question

(a) What magnitude point charge creates a 10,000 NIC elec- tric field at a distance of 0.250 m? (b) How large is the field at 10.0 m?
Please show work and formulas used for credit

Answer 1

(a) The magnitude of electric field of a point charge at of charge a distance from it is given by

$E=\frac{1}{4\pi\epsilon_o}\frac{q}{r^2}=9\times10^9\frac{q}{r^2}$

Since we have to find for a given and , we rearrange the above formula to make the subject. Hence we get

$q=\frac{E\times r^2}{9\times10^9}=\frac{10000\times0.25^2}{9\times10^9}=6.944\times10^{-8}~\mathrm{C}$

(b) The electric field at the distance $r=10~\mathrm{m}$ is given by

$E=\frac{1}{4\pi\epsilon_o}\frac{q}{r^2}=\frac{9\times10^9\times6.944\times10^{-8}}{10^2}=6.2496~\mathrm{N/C}$

Answer 2

(a) The magnitude of electric field of a point charge at of charge a distance from it is given by

$E=\frac{1}{4\pi\epsilon_o}\frac{q}{r^2}=9\times10^9\frac{q}{r^2}$

Since we have to find for a given and , we rearrange the above formula to make the subject. Hence we get

$q=\frac{E\times r^2}{9\times10^9}=\frac{10000\times0.25^2}{9\times10^9}=6.944\times10^{-8}~\mathrm{C}$

(b) The electric field at the distance $r=10~\mathrm{m}$ is given by

$E=\frac{1}{4\pi\epsilon_o}\frac{q}{r^2}=\frac{9\times10^9\times6.944\times10^{-8}}{10^2}=6.2496~\mathrm{N/C}$

Answer 3

(a) The magnitude of electric field of a point charge at of charge a distance from it is given by

$E=\frac{1}{4\pi\epsilon_o}\frac{q}{r^2}=9\times10^9\frac{q}{r^2}$

Since we have to find for a given and , we rearrange the above formula to make the subject. Hence we get

$q=\frac{E\times r^2}{9\times10^9}=\frac{10000\times0.25^2}{9\times10^9}=6.944\times10^{-8}~\mathrm{C}$

(b) The electric field at the distance $r=10~\mathrm{m}$ is given by

$E=\frac{1}{4\pi\epsilon_o}\frac{q}{r^2}=\frac{9\times10^9\times6.944\times10^{-8}}{10^2}=6.2496~\mathrm{N/C}$