Question

The strength of an electric field at 5.0 cm from a point charge is 100.0 N/C. What is the magnitude of the source charge?

Answer 1

here we can apply this formula:
\[\Large E = K\frac{Q}{{{d^2}}}\]
where \(E=100\), \(K=9 \cdot 10^9\), and \(d=0.05\) meters

Answer 2

So the equation would be \[100=9\times10^9 \frac{ Q }{ (0.05^2) }\] ?

Answer 3

that's right!

Answer 4

here is the next step:
\[\huge Q = \frac{{E{d^2}}}{K} = \frac{{{{10}^2} \cdot 25 \cdot {{10}^{ - 4}}}}{{9 \cdot {{10}^9}}} = ...?\]

Answer 5

So then
\[\frac{ .25 }{ 9\times10^9 }=2.7^-11\]?

Answer 6

This is the result that I got in the end:
\[Q=2.7\times10^-11C\]
Is that correct?

Answer 7

yes! It is correct!

Answer 8

Thank you! (: