First slide

Electric field

Question

A positive point charge 50 $μ$ C is located in the plane xy at a point with radius vector ${\vec{r}}_{0} = 2 \hat{i} + 3 \hat{j}$ . The electric field vector, $\vec{E}$ at a point with radius vector $\vec{r} = 8 \hat{i} - 5 \hat{j}$ , where r₀ and r are expressed in meter, is

Easy

A

$(1.4 \hat{i} - 2.6 \hat{j}) {kNC}^{- 1}$
B

$(1.4 \hat{i} + 2.6 \hat{j}) k {NC}^{- 1}$
C

$(2.7 \hat{i} - 3.6 \hat{j}) {kNC}^{- 1}$
D

$(2.7 \hat{i} + 3.6 \hat{j}) {kNC}^{- 1}$

Solution

$\begin{array}{l} As \vec{E} = \frac{1}{4 π ε_{0}} \frac{q}{{|\vec{r} - {\vec{r}}_{0}|}^{3}} (\vec{r} - {\vec{r}}_{0}) \\ \vec{M N} = \vec{r} - {\vec{r}}_{0} = (\hat{i} - 5 \hat{j}) - (2 \hat{i} + 3 \hat{j}) = (6 \hat{i} - 8 \hat{j}) \end{array}$
$\begin{array}{l} |\vec{r} - {\vec{r}}_{0}| = \sqrt{6^{2} + 8^{2}} = 10 m \\ \vec{E} = 9 \times 10^{9} \times \frac{50 \times 10^{- 6}}{(10)^{3}} (6 \hat{i} - 8 \hat{j}) \\ \vec{E} = (2.7 \hat{i} - 3.6 \hat{j}) k {NC}^{- 1} \end{array}$

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