At the origin, there is an A+10 nC charge. What is the electric field's strength at the coordinates (x,y)=(5.0 cm,0 cm)?

The field's strength at point (x,y), = (5 cm, 0 cm). Since there is a distance between the charge and that location.r=x2+y2&nbsp;=(0.05)2+(0)2&nbsp;=0.05&nbsp;m&nbsp;E=kqr2&nbsp;=(9×109)(9×10−9)(0.052)&nbsp;=3.6×104&nbsp;N/C

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jubateee

Answered question

2021-12-12

At the origin, there is an A+10 nC charge. What is the electric field's strength at the coordinates (x,y)=(5.0 cm,0 cm)?

Answer & Explanation

Becky Harrison

Beginner2021-12-13Added 40 answers

The field's strength at point (x,y), = (5 cm, 0 cm). Since there is a distance between the charge and that location.
$r = \sqrt{x^{2} + y^{2}}$
$= \sqrt{{(0.05)}^{2} + {(0)}^{2}}$
$= 0.05 m$
$E = \frac{k q}{r^{2}}$
$= \frac{(9 \times 10^{9}) (9 \times 10^{- 9})}{({0.05}^{2})}$
$= 3.6 \times 10^{4}$ N/C

Melinda McCombs

Beginner2021-12-14Added 38 answers

I need the same, but for (x, y) = (-5cm, -5 cm)

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