Question

Compute the electric field experienced by a negative test charge given the following conditions in a vacuum: b. Source charge: +6.78 × 10^–15 C Distance from source charge: 4.6 × 10^–19 m (Your answer must be in this format : 1.11x10^1 N/C, -1.11x10^-1 N/C) 

Answer 1

$$E=k\frac{q}{r^2}\\ ={\rm 9*10^9\: N\cdot m^2/C^2*\frac{6.78 × 10^{–15}\: C }{(4.6 × 10^{–19}\: m)^2}}\\ =2.88*10^{32}\:\rm V/m$$