Respuesta :
Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, [tex]\lambda = 1\ nC = 1\times 10^{- 9}\ C[/tex]
Charge density of rod 2, [tex]\lambda = - 1\ nC = - 1\times 10^{- 9}\ C[/tex]
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:
[tex]\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}[/tex]
Also,
[tex]\vec{E} = \frac{2K\lambda }{R}[/tex] (1)
where
K = electrostatic constant = [tex]\frac{1}{4\pi \epsilon_{o} R}[/tex]
R = Distance
[tex]\lambda[/tex] = 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):
[tex]\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C[/tex]
[tex]\vec{E} = 1800\ N/C[/tex] (towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):
[tex]\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C[/tex]
[tex]\vec{E'} = 1800\ N/C[/tex] (towards)
Now, the total field at the origin is the sum of both the fields:
[tex]\vec{E_{net}} = 1800 + 1800 = 3600\ N/C[/tex]
