Physics, 24.12.2019 05:31, ashgold324

Along, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nc/m . a second long, thin rod parallel to the z-axis is located at x = +1 cm and carries a uniform negative charge density λ = - 1 nc/m. what is the electric field at the origin?

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answered: dbethel8408

The electric field at origin is 3600 N/C

Solution:

As per the question:

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

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

$\lambda$ = linear charge density

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):

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

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

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

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

answered: Guest

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explanation:

answered: Guest

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