Answer:
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Answer:

vector with direction equal to the axis X.
Explanation:
We use the Gauss Law and the superposition law in order to solve this problem.
<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

Then:

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula 

E = 1461.95 N/C
c) The electric field E is calculated as:

E = 239.76 N/C
Wee can use here kinematics
as we know that

for shorter tree we know that


now since we know that other tree is twice high
So height of other tree is y = 39.2 m
now again by above equation



so the time taken is 2.83 s
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