Question

What is the magnitude of the electric field produced by a charge of magnitude 5.00 μC at a distance of 1.00 m?

Answer 1

Answer:

magnitude of electric field  E = 45  × 10³ N/C

Explanation:

given data

magnitude of the charge is q = 5.00 μC = 5 × $10^{-6}$  C

distance = 1 m

to find out

magnitude of the electric field

solution

we will apply here magnitude of electric field formula that is

magnitude of electric field  E = k × $\frac{q}{r^2}$     .......................1

put here value we get

magnitude of electric field  E = 9 × $10^{9}$  × $\frac{5*10^{-6}}{1^2}$

magnitude of electric field  E = 45  × 10³ N/C