Question

Calculate the electric field 0.245 cm from a point charge of 25.6 nC.

Answer 1

The electric field of a point charge is given as follows:

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

where $k = 9\times 10^9N\cdot m^2/C^2$, $q = 25.6\times10^{-9}C$, and $r = 0.245cm = 0.00245m$. Thus, obtain:

$$E = 9\times 10^9\cdot \dfrac{25.6\times 10^{-9}}{0.00245^2} \approx 3.84\times 10^7V/m$$

Answer. $3.84\times 10^7V/m$.