Question

Calculate the electric field experienced by a negative test charge given that source charge is +6.78 𝑥 10−15𝐶 and 7.9 𝑥 10−19𝑚 as the distance from the source charge

Answer 1

Electric field

$E=\frac{Kq}{r^2}$

$E=\frac{9\times10^9\times6.78\times10^{-15}}{(7.9\times10^{-19})^2}$

$E=9.77\times10^{31}N/c$