خطا
میدان الکتریکی برآیند را در نقطهٔ A محاسبه میکنیم: $\begin{align} & {{\overrightarrow{E}}_{A}}={{\overrightarrow{E}}_{1}}+{{\overrightarrow{E}}_{2}} \\ & {{E}_{A}}={{E}_{1}}+{{E}_{2}} \\\end{align}$ $\begin{align} & {{E}_{A}}=\frac{k\left| {{q}_{2}} \right|}{r_{2}^{2}}-\frac{k\left| {{q}_{1}} \right|}{r_{1}^{2}}=9\times {{10}^{9}}(\frac{5\times {{10}^{-9}}}{25\times {{10}^{-4}}}-\frac{5\times {{10}^{-9}}}{225\times {{10}^{-4}}})=9\times (\frac{45-5}{225})\times {{10}^{4}}=9\times \frac{40\times {{10}^{4}}}{225}=1/6\times {{10}^{4}}\frac{N}{C} \\ & F=\left| q \right|E\Rightarrow 8\times {{10}^{-5}}=\left| q \right|\times 1/6\times {{10}^{4}}\Rightarrow \left| q \right|=\frac{8\times {{10}^{-5}}}{1/6\times {{10}^{4}}}=5\times {{10}^{-9}}=5nC \\\end{align}$