Question

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

Answer 1

Step-by-step 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 * {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 * {{10}^9}} \right)\left( {10\;{\rm{nC}}} \right)}}{{{{\left( {5\sqrt 2 \;{\rm{cm}}} \right)}^2}}}\\\\ = \frac{{\left( {9 * {{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 * {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 * {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 * {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 * {{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