Answer
(a) $6.74\times 10^4\frac{N}{C}$
(b) $1.69\times 10^4\frac{N}{C}$
Work Step by Step
(a) We can find the magnitude of electric field as follows:
$E=K\frac{q}{r^2}$
We plug in the known values to obtain:
$E=8.99\times 10^9\times \frac{7.50\times 10^{-6}}{(1)^2}$
$E=6.74\times 10^4\frac{N}{C}$
(b) We can find the magnitude of electric field as follows:
$E=K\frac{q}{r^2}$
We plug in the known values to obtain:
$E=8.99\times 10^9\times \frac{7.50\times 10^{-6}}{(2.00)^2}$
$E=1.69\times 10^4\frac{N}{C}$
