## Physics (10th Edition)

Let the point charge be q. Then: Electric field at a distance $r_{1}$ $(E_{1})=\frac{1}{4\pi\epsilon_0}\frac{q}{r_{1}^{2}}$= 248 N/C and electric field at a distance $r_{2} (E_{2})$=$\frac{1}{4\pi\epsilon_{0}}\frac{q}{r_{2}^{2}}$= 132 N/C ⇒$\frac{E_{1}}{E_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}= \frac{248 N/C}{132 N/C}= 1.88$ ⇒$\frac{r_{2}}{r_{1}}=\sqrt {1.88} = 1.37$