Question

2. Determine the magnitude of electric field value if a distance of 35 cm from a 0.00050 C

charge.

Answer 1

The electric field due to the point charge

$$E=k\frac{q}{r^2}$$

$$E=9*10^9\frac{0.00050}{0.35^2}=3.7\times 10^7\:\rm N/C$$