Question

What is the electric field at the position (x1,y1)=(−5.0 cm, 5.0 cm) in component form?

Answer 1

Answer:

Explanation:

Calculate the distance between origin O $\left( {0\;{\rm{cm,}}\;0\;{\rm{cm}}} \right)$ and the point $\left( { - 5.0\;{\rm{cm,}}\;5.0\;{\rm{cm}}} \right)$ as follows:

Substitute 0 cm for ${x_1} ,0cm$ for ${y_1} , - 5.0\;{\rm{cm}}$ for ${x_2}$ , and $5.0\;{\rm{cm}}$ for ${y_2}$  in $r = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

$\begin{array}{c}\\r = \sqrt {{{\left( { - 5.0\;{\rm{cm}}\; - 0\;{\rm{cm}}} \right)}^2} + {{\left( {5.0\;{\rm{cm}}\; - 0\;{\rm{cm}}} \right)}^2}} \\\\ = 5\sqrt 2 \;{\rm{cm}}\\\end{array}$

The electric field is,

$E = k\frac{q}{{{r^2}}}$

Substitute $9 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}$ for k, 10 nC for q, and $5\sqrt 2 \;{\rm{cm}}$ for r in $E = k\frac{q}{{{r^2}}}$

$\begin{array}{c}\\E = \frac{{\left( {9 \times {{10}^9}} \right)\left( {10\;{\rm{nC}}} \right)}}{{{{\left( {5\sqrt 2 \;{\rm{cm}}} \right)}^2}}}\\\\ = \frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {10\;{\rm{nC}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{C}}}}{{1\;{\rm{nC}}}}} \right)}}{{{{\left( {\left( {5\sqrt 2 \;{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)} \right)}^2}}}\\\\ = 1.8 \times {10^4}\;{\rm{N/C}}\\\end{array}$

 

Therefore, the electric field at the position $\left( { - 5.0\;{\rm{cm,}}\;5.0\;{\rm{cm}}} \right)$ is $1.8 \times {10^4}\;{\rm{N/C}}$

The electric field is inversely proportional to the square of the distance r.

The electric field in vector form is,

$\vec E = {\vec E_{2x}}i + {\vec E_{2y}}j$

Here, ${\vec E_{2x}}$ is the x component of electric field vector, and ${\vec E_{2y}}$ is the y component of electric field vector.

The x component of electric field vector is $- E\cos \theta$, and the y component of electric field vector is $E\sin \theta$.

Substitute $- E\cos \theta$ for ${\vec E_{2x}}$, and $E\sin \theta$ for ${\vec E_{2y}}$ in $\vec E = {\vec E_{2x}}i + {\vec E_{2y}}j$

$\begin{array}{c}\\\vec E = \left( { - E\cos \theta } \right)i + \left( {E\sin \theta } \right)j\\\\ = \left( E \right)\left( {\left( { - \cos \theta } \right)i + \left( {\sin \theta } \right)j} \right)\\\end{array}$

Substitute $1.8 \times {10^4}\;{\rm{N/C}}$ for E, and 450 for $\thetaθ$ in $\vec E = \left( E \right)\left( {\left( { - \cos \theta } \right)i + \left( {\sin \theta } \right)j} \right)$  

$\begin{array}{c}\\\vec E = \left( {1.8 \times {{10}^4}\;{\rm{N/C}}} \right)\left( { - \cos 45i + \sin 45j} \right)\\\\ = - \left( {1.27\;{\rm{N/C}}} \right)i + \left( {1.27\;{\rm{N/C}}} \right)j\\\end{array}$

Therefore, the electric field vector in component form is,

$\left( {{{\vec E}_{2x}},{{\vec E}_{2y}}} \right) = \left( { - 1.27\;{\rm{N/C}},\,1.27\;{\rm{N/C}}} \right){\rm{N/C}}$

The point $\left( { - 5.0\;{\rm{cm,}}\;5.0\;{\rm{cm}}} \right)$ lies in the second quadrant, so the x component of electric field vector is negative, and y component of the electric field vector is positive