in the diagram, q1 = +7.50 μc and q2 = -4.25 μc. what is the electric field at point p? include a + or…

Explanation:

Step1: Convert distances to SI units

$r_1=(4.24 + 1.20)\times10^{- 2}\text{ m}=5.44\times10^{-2}\text{ m}$, $r_2 = 1.20\times10^{-2}\text{ m}$

Step2: Calculate electric - field due to $q_1$

The electric - field formula is $E=\frac{kq}{r^{2}}$, where $k = 9\times10^{9}\text{ N}\cdot\text{m}^{2}/\text{C}^{2}$. For $q_1=7.50\times10^{-6}\text{ C}$, $E_1=\frac{kq_1}{r_1^{2}}=\frac{9\times10^{9}\times7.50\times10^{-6}}{(5.44\times10^{-2})^{2}}$ $E_1=\frac{67.5\times10^{3}}{2.96\times10^{-3}}\text{ N/C}\approx2.28\times10^{7}\text{ N/C}$ (directed away from $q_1$ towards point $P$)

Step3: Calculate electric - field due to $q_2$

For $q_2=-4.25\times10^{-6}\text{ C}$, $E_2=\frac{k|q_2|}{r_2^{2}}=\frac{9\times10^{9}\times4.25\times10^{-6}}{(1.20\times10^{-2})^{2}}$ $E_2=\frac{38.25\times10^{3}}{1.44\times10^{-4}}\text{ N/C}\approx2.66\times10^{8}\text{ N/C}$ (directed towards $q_2$ from point $P$)

Step4: Calculate the net electric - field

Since the electric fields $E_1$ and $E_2$ are in the same direction (towards the negative charge $q_2$), $E = E_1+E_2$. $E=(2.28\times10^{7}+2.66\times10^{8})\text{ N/C}=(0.228\times10^{8}+2.66\times10^{8})\text{ N/C}=2.89\times10^{8}\text{ N/C}$

Answer:

$2.89\times10^{8}$