Question

A point charge q=-8nc is located at the origin. Find the electric field vector

Answer 1

The electric field vector is $-10.8i+14.4j$

Explanation:

Given that,

Charge q=-8nC

Suppose find the electric field vector at the point x = 1.2 m, y = -1.6 m

We need to calculate the electric field vector

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}\hat{r}$...(I)

We need to calculate the value of $\hat{r}$

$\hat{r}=\dfrac{\vec{r}}{r}$

Put the value into the formula

$\hat{r}=\dfrac{(1.2-0)i+(-1.6-0)j}{\sqrt{(1.2)^2+(1.6)^2}}$

$\hat{r}=\dfrac{1.2i-1.6j}{2}$

Put the value of $\hat{r}$ in equation (I)

$E=9\times10^{9}\times\dfrac{-8\times10^{-9}}{2^2}\timesdfrac{1.2i-1.6j}{2}$

$E=-10.8i+14.4j$

Hence, The electric field vector is $-10.8i+14.4j$

Learn more :

Topic : Electric field

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