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What is the electric field at a point 0.450 m to the left of a -5.77*10^-9 C charge?
Include a + or - sign to indicate the direction the field.

The answer is -0.13


Sagot :

The answer is -0.13
Hope this helps

256.2 is the answer.

Here, we are asked to calulate the electric field.

Using the following formula:

E=k*lql/r^2

Yeah! Something that can easily bamboozle you.

So:

E=electric field

k=constant (8.99x10^9)

q=charge

ll= absolute value marks ( I still remember from my Algebra Course))

r=distance

Now, let's plug in the numbers.

E=8.99x10^9*l-5.77x10^-9l/(0.450)^2

And  yes! That's a LOT!

But I solved the same problem on paper 2 minutes ago:)

If you perform the calculations, you will get 256.2

Well. THIS one was correct or Acellus.

So our answer is:

256.2 N/C

Newtons per Coulomb

Hope this helps:)