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Katyanochek1 [597]

1 year ago

7

A very long, straight wire has charge per unit length 3.20 x 10^-10 c>m. At what distance from the wire is the electric-field

magnitude equal to 2.50 nc?

1 answer:

vovangra [49]1 year ago

5 0

at 2.304×10⁹m the electric field magnitude equal to 2.50 nc.

electric field due to a long uniformly charged wire is given as,

E= 2kλ/r

where,

k is a constant

λ is uniform charge density

r is the distance from the wire

λ = 3.20 × 10⁻¹⁰C/m

k= 9×10⁹

therefore r = 2kλ/E

on substituting value we get,

r= 2.304×10⁹m

learn more about electric field due to a uniformly charged wire here:

brainly.com/question/28166144

#SPJ4

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2 years ago

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Answer:

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As per the question:

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Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

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Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

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Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

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