Question

The strength of an electric field at 2.4 m from a point charge is 250.0 N/C. What is the magnitude of the source charge?

**How can I show the work for this? Thank you :)

Answer 1

here we have to apply this formula:
\[E = K\frac{Q}{{{r^2}}}\]
where E is the electric field, Q is the electric charge, and r is the distance from the electric charge to the point of observation

Answer 2

as usually K = 9*10^9 N*m^2/Coulomb^2

Answer 3

so we have to write the inverse formula, to find Q. What is that formula?

Answer 4

ohh okay! and i'm not sure :( i forgot :(

Answer 5

here is it:
\[\Large Q = \frac{{E{r^2}}}{K}\]

Answer 6

so substituting your data we get:
\[\Large Q = \frac{{E{r^2}}}{K} = \frac{{250 \times {{\left( {2.4} \right)}^2}}}{{9 \times {{10}^9}}} = ...?\]

Answer 7

ahh okay, that's right!! :)

so then 1440/9*10^9 ?

i'm not sure what that gets? My calculator says 1.6 E^-7.. is that correct?

Answer 8

ok! we have:
\[\large Q = \frac{{E{r^2}}}{K} = \frac{{250 \times {{\left( {2.4} \right)}^2}}}{{9 \times {{10}^9}}} = 1.6 \times {10^{ - 7}} = 0.16 \times {10^{ - 6}} = 0.16\mu C\]

Answer 9

ohhh so you condensed it! i see:)

yay!! so what happens next? is there more? :O or is 0.16µC the final answer?

Answer 10

It is the final answer!

Answer 11

yay!! thank you:)

Answer 12

thank you! :)