# Chapter 23 - Gauss' Law - Problems - Page 683: 62a

$E=4.0\times 10^6\frac{N}{C}$

#### Work Step by Step

The direction of electric field at $P_1$ is away from $q_1$ and we can find its magnitude as: $E=\frac{1 }{4\pi \epsilon_{\circ}}\frac{q}{r^2}$ We plug in the known values to obain: $E=8.99\times 10^9\times \frac{1.0\times 10^{-7}}{(0.015)^2}=4.0\times 10^6\frac{N}{C}$

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