Question

An electron is fixed at the origin as shown. What is the electric field at A?

• A. $3.6\times {10}^{-19}\text{N/C}\hat{i}$
• B. $-3.6\times {10}^{-10}\text{N/C}\hat{i}$
• C. $3.6\times {10}^{-10}\text{N/C}\hat{i}$
• D. None of these

Answer 1

The correct option is B $-3.6\times {10}^{-10}\text{N/C}\hat{i}$

To calculate the Electric field, we need to know the charge. We know electrons are negatively chargedparticles and one electron has a charge of –1.6 × 10^–19 C. Since it is negatively charged its electric field will point towards itself (that is along $\overrightarrow{l}$ )

The second thing we need to know is the distance of charge from the point (where we want to find the Field).

$|\overrightarrow{E}|=\frac{1}{4\pi {\epsilon }_0}\frac{|Q|}{r^2}\Rightarrow |\overrightarrow{E}|=\frac{1}{4\pi {\epsilon }_0}\times \frac{1.6\times {10}^{19}}{\left(2^2\right)}\approx 3.6\times {10}^{-10}N/C(-\hat{J})So,\overrightarrow{E}=-3.6\times {10}^{-10}N/C\hat{J}$