# Consider point a which is 62 cm north of a point charge Q = -3.8 \space \mu C and point b which...

## Question:

Consider point {eq}a {/eq} which is 62 cm north of a point charge {eq}Q = -3.8 \space \mu C {/eq} and point {eq}b {/eq} which is 88 cm west of {eq}Q {/eq}.

a. Determine {eq}V_b - V_a {/eq}.

b. Determine the magnitude of {eq}\vec{E}_b - \vec{E}_a {/eq}.

c. Determine the direction of {eq}\vec{E}_b - \vec{E}_a {/eq}.

## Electric Field and Potential due to Point Charge:

The electric field at a point {eq}A {/eq} due to a point charge {eq}Q {/eq} having charge {eq}q\ C {/eq} such that {eq}\overrightarrow{QA} = \overrightarrow{r} {/eq} is, $$\overrightarrow{E} = \frac{q \overrightarrow{r}}{4\pi\epsilon_0 |\overrightarrow{r}|^3}\ N/C$$

The electric potential at point A is, $$V = \frac{q}{4\pi\epsilon_0 |\overrightarrow{r}|}\ V$$

Let origin be at the charge itself. Let North direction be +y axis and East direction be +x axis.

Then, $$\overrightarrow{QA} = 0.62j\ m\\ \overrightarrow{QB} = -0.88i\ m$$

Therefore, electric potential at these point will be, $$V_a = \frac{q}{4\pi\epsilon_0 |\overrightarrow{r}|}\ V = \frac{-3.8\times 10^{-6}}{4\pi\epsilon_0 \cdot 0.62}\ V = -55.084\ kV\\ V_b = \frac{-3.8\times 10^{-6}}{4\pi\epsilon_0 \cdot 0.88}\ V = -38.809\ kV\\ V_b-V_a = 16.275\ kV$$

The electric field at these point is given by, $$\overrightarrow{E}_a = \frac{q \overrightarrow{r}}{4\pi\epsilon_0 |\overrightarrow{r}|^3}\ N/C = \frac{-3.8\times 10^{-6} \cdot 0.62j}{4\pi\epsilon_0 |0.62j|^3}\ N/C = -88.846j\ kN/C\\ \overrightarrow{E}_b = \frac{-3.8\times 10^{-6} \cdot (-0.88i)}{4\pi\epsilon_0 |-0.88i|^3}\ N/C = 44.102i\ kN/C\\ \overrightarrow{E}_b - \overrightarrow{E}_a = (44.102i+88.846j)\ kN/C\\ |\overrightarrow{E}_b - \overrightarrow{E}_a| = \sqrt{44.102^2+88.846^2} = 99.19\ kN/C\\ \tan \theta = \frac{88.846}{44.102} = 2.015\\ \theta = 63.61^{\circ}$$

These are the magnitude and direction.