Answer to Question #144251 in Electric Circuits for gsk

Question

it shows the charge distribution within a thundercloud.There is a charge of 40 C at a height of 10 km,−40 C at 5 km,and 10 C at 2 km. Considering these charges to be pointlike,find the net electric field produced at a height of 8 km and ahorizontal distance of 3 km.

Answer 1

Explanations & Calculations

• Electric field is the electric force generated on a positive test unit charge (+1C) by the other charges.
• And the electric field (E) at a point at a distance r by a charge q is $\small E = \large \frac{kq}{r^2}$ (NC^-1)
• If this r could be written by Pythagoras relationship as $\small r^2=x^2+y^2$ then,

$\qquad\qquad \begin{aligned} \small E &= \small \frac{kq}{r^2}=\frac{kq}{x^2+y^2}=\frac{kq}{x^2}+\frac{kq}{y^2} \end{aligned}$

• E could be written in its orthogonal components in such a situation.
• This is the easiest way to find the net electric field at the given point as the point is shifted right from the main charge distribution & the distances from each charge to the given point follow Pythagoras pattern.

• Sign convention: Downward & the leftward directions are negative & the upward & the rightward are positive.

• Field due to the +40C charge

$\qquad\qquad E_1= \begin{cases} \begin{aligned} \small y_1=\frac{k\,40C}{(10-8)^2}=\frac{-40k}{4} :\text{vertical} \end{aligned}\\ \begin{aligned} \small x_1= \frac{k\,40C}{(3)^2}=\frac{+40k}{9}: \text{horizontal} \end{aligned}\\ \end{cases}$

• Field due to the -40C charge

$\qquad\qquad E_2= \begin{cases} \begin{aligned} \small y_2=\frac{k\,40C}{(5-8)^2}=\frac{-40k}{9} :\text{vertical} \end{aligned}\\ \begin{aligned} \small x_2= \frac{k\,40C}{(3)^2}=\frac{-40k}{9}: \text{horizontal} \end{aligned}\\ \end{cases}$

• Field due to the +10C charge

$\qquad\qquad E_3= \begin{cases} \begin{aligned} \small y_3=\frac{k\,10C}{(2-8)^2}=\frac{+10k}{36} :\text{vertical} \end{aligned}\\ \begin{aligned} \small x_3= \frac{k\,10C}{(3)^2}=\frac{+10k}{9}: \text{horizontal} \end{aligned}\\ \end{cases}$

• Then the net force can be calculated by summing up both the vertical & horizontal components. Therefore, the net force in component form is

$\qquad\qquad E_{net}= \begin{cases} \begin{aligned} \small Y_{net}=\frac{(-40k)}{4}+ \frac{(-40k)}{9}+\frac{10k}{36}=\frac{(-85k)}{6}:\text{verticle} \end{aligned}\\ \begin{aligned} \small X_{net}= \frac{40k}{9}+\frac{(-40)k}{9}+\frac{10k}{9}=\frac{10k}{9}: \text{horizontal} \end{aligned}\\ \end{cases}$ or $\small E_{net}=\frac{-85k}{6}i+\frac{10k}{9}j$

$\qquad\qquad \begin{aligned} \small |E_{net}| &= \small \sqrt{(\frac{85k}{6})^2+(\frac{10k}{9})^2}\\ &= \small 14.210k\\ &= \small 14.210\times (8.988\times 10^9Nm^2/C^2)\\ &= \small \bold{1.277\times 10^{11} N/C} \end{aligned}$