Question

Two charges (+3µC on the left of the -5µC) are separated by a distance of 5um on a horizontal line. Point P is located 10µm above the 3µC charge.

a) Draw a sketch of the setting discussed in the question above. Make sure to label all points discussed. (2marks)

b) Calculate the field at point P (3marks)

c) Calculate the potential at point P 

(1 mark)

Answer 1

(a) Let $q_1=+3\ \mu C$, $q_2 = -5\ \mu C$. Let the distance between charges $q_1$ and $q_2$ be $r_{12}=5\ \mu m$, the distance from the charge $q_1$ to the point P above it be $r_{1P}=10\ \mu m$. Let's draw a sketch:

(b) The electric field due to point charge $q_1$ at point P directed upward (away from the charge $q_1$), the electric field due to point charge $q_2$ at point P directed toward charge $q_2$.

Let’s first find the magnitudes of the electric fields $E_1$ and $E_2$:

$$E_1=\dfrac{k|q_1|}{r_{1P}^2}=\dfrac{9.0\cdot10^{9}\ \dfrac{N\cdot m^2}{C^2}\cdot|3\cdot10^{-6}\ C|}{(10\cdot10^{-6}\ m)^2}=2.7\cdot10^{14}\ \dfrac{N}{C},$$ $$E_2=\dfrac{k|q_2|}{r_{2P}^2}=\dfrac{9.0\cdot10^{9}\ \dfrac{N\cdot m^2}{C^2}\cdot|-5\cdot10^{-6}\ C|}{\sqrt{(10\cdot10^{-6}\ m)^2+(5\cdot10^{-6}\ m)^2}}=4.0\cdot10^{9}\ \dfrac{N}{C}.$$

Let’s find projections of the fields $E_1$ and $E_2$ on axis $x$- and $y$:

$$E_{1x}=0, E_{1y}=E_1=2.7\cdot10^{14}\ \dfrac{N}{C},$$ $$E_{2x}=E_2cos\theta=E_2\dfrac{r_{1P}}{r_{2P}},$$ $$E_{2x}=4.0\cdot10^{9}\ \dfrac{N}{C}\cdot\dfrac{10\cdot10^{-6}\ m}{\sqrt{(10\cdot10^{-6}\ m)^2+(5\cdot10^{-6}\ m)^2}}=3.6\cdot10^9\ \dfrac{N}{C},$$ $$E_{2y}=E_2sin\theta=E_2\dfrac{r_{12}}{r_{2P}},$$ $$E_{2y}=4.0\cdot10^{9}\ \dfrac{N}{C}\cdot\dfrac{5\cdot10^{-6}\ m}{\sqrt{(10\cdot10^{-6}\ m)^2+(5\cdot10^{-6}\ m)^2}}=1.78\cdot10^9\ \dfrac{N}{C}.$$

Then, we can find $x$- and $y$-components of the net electric field at point P:

$$E_x=E_{2x}=3.6\cdot10^9\ \dfrac{N}{C},$$ $$E_y=E_{1y}-E_{2y}=2.7\cdot10^{14}\ \dfrac{N}{C}-1.78\cdot10^9\ \dfrac{N}{C}=2.69\cdot10^{14}\ \dfrac{N}{C}.$$

We can find the net electric field at point P from the Pythagorean theorem:

$E_{net}=\sqrt{E_x^2+E_y^2}=\sqrt{(3.6\cdot10^9\ \dfrac{N}{C})^2+(2.69\cdot10^{14}\ \dfrac{N}{C})^2}=2.69\cdot10^{14}\ \dfrac{N}{C}.$

We can find the direction of the net electric field from the geometry:

$$\theta=tan^{-1}(\dfrac{E_y}{E_x})=tan^{-1}(\dfrac{2.69\cdot10^{14}\ \dfrac{N}{C}}{3.6\cdot10^9\ \dfrac{N}{C}})=89.9^{\circ}.$$

(c) We can find the potential at point P as follows:

$$V=V_1+V_2=\dfrac{kq_1}{r_{1P}}+\dfrac{kq_2}{r_{2P}},$$ $$V=\dfrac{9.0\cdot10^{9}\ \dfrac{N\cdot m^2}{C^2}\cdot3\cdot10^{-6}\ C}{10\cdot10^{-6}\ m}+\dfrac{9.0\cdot10^{9}\ \dfrac{N\cdot m^2}{C^2}\cdot(-5\cdot10^{-6}\ C)}{\sqrt{(10\cdot10^{-6}\ m)^2+(5\cdot10^{-6}\ m)^2}},$$ $$V=-1.32\cdot10^{9}\ V.$$