Problem Given EEE, find rrr. Find the distance from a point charge +5 μC+5\ \mu\text{C}+5 μC at which the magnitude of the electric field is ∣E⃗∣=100 V/m\abs{\vv{E}} = 100\ \text{V/m}E=100 V/m. Solution Symbolic setup. The field of a point charge is E=k qr2E = k\,\dfrac{q}{r^2}E=kr2q. Isolating rrr: r=k qEr = \sqrt{\frac{k\,q}{E}}r=Ekq Numerical substitution. With k=8,99⋅109 N\cdotpm2/C2k = 8{,}99\cdot 10^9\ \text{N·m}^2/\text{C}^2k=8,99⋅109 N\cdotpm2/C2, q=5⋅10−6 Cq = 5\cdot 10^{-6}\ \text{C}q=5⋅10−6 C, E=100 V/mE = 100\ \text{V/m}E=100 V/m: r=8,99⋅109⋅5⋅10−6100=4,495⋅104100=449,5≈21,2 mr = \sqrt{\frac{8{,}99\cdot 10^9 \cdot 5\cdot 10^{-6}}{100}} = \sqrt{\frac{4{,}495\cdot 10^4}{100}} = \sqrt{449{,}5} \approx 21{,}2\ \text{m}r=1008,99⋅109⋅5⋅10−6=1004,495⋅104=449,5≈21,2 m r≈21,2 m\ev{r \approx 21{,}2\ \text{m}}r≈21,2 m Links Topics: Electric field and potential Concepts: Electric field Skills: Symbolic setup