Get the solution to "A+10 nC charge is located at the origin...."

Oberlaudacu

Answered question

2022-01-15

A+10 nC charge is located at the origin. What is the strength of the electric field at the position (x,y)=(−5.0 cm,−5.0 cm)

Pademagk71

Beginner2022-01-16Added 34 answers

The strength of the field at the point, (x, y) = (-5cm, -5 cm)
Since, the distance from the charge to that point is,

r = \sqrt{x^{2} + y^{2}}

= \sqrt{{(- 0.05)}^{2} + {(- 0.05)}^{2}}

= 0.071 m

Let the electric field strenght at the above point is

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

= \frac{(9 \times 10^{9}) (9 \times 10^{- 9})}{({0.071}^{2})}

= 1.8 \times 10^{4}

N/C

usumbiix

Beginner2022-01-17Added 33 answers

I think that answer must be negative

alenahelenash

Expert2022-01-23Added 556 answers

Original answer is correct

alenahelenash

Expert2022-01-23Added 556 answers

No, you are wrong, answer is correct

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