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dimaraw [331]
3 years ago
5

1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.​

Physics
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:

r = 20.22 m

Explanation:

Given that,

Charge,q=2.5\times 10^{-6}\ C

Electric field, E=55\ N/C

We need to find the distance. We know that, the electric field a distance r is as follows :

E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m

So, the required distance is 20.22 m.

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<em>Hi</em><em>,</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>Energy</em><em> </em><em>is</em><em> </em><em>defined</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>capacity</em><em> </em><em>or</em><em> </em><em>ability</em><em> </em><em>to</em><em> </em><em>do</em><em> </em><em>work</em><em>.</em><em> </em><em>It's</em><em> </em><em>SI</em><em> </em><em>unit</em><em> </em><em>is</em><em> </em><em>Joule</em><em>.</em>

<em>here</em><em>,</em>

<em>Joule</em><em> </em><em>=</em><em> </em><em>(</em><em>kg</em><em>×</em><em>m</em><em>×</em><em>m</em><em>)</em><em>/</em><em>(</em><em>s</em><em>×</em><em>s</em><em>)</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>kg</em><em>×</em><em>m</em><em>^</em><em>2</em><em>/</em><em>s</em><em>^</em><em>2</em><em>.</em>

<em>Therefore</em><em>, </em><em> </em><em>the</em><em> </em><em>derived</em><em> </em><em>unit</em><em> </em><em>is</em><em> </em><em>kg</em><em>.</em><em>m</em><em>^</em><em>2</em><em> </em><em>by</em><em> </em><em>s</em><em>^</em><em>2</em><em>.</em>

<em>Hope it helps</em><em>.</em><em>.</em><em>.</em>

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