(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.
(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.
<h3>Electric field at a point on the Gaussian surface</h3>
The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;
E = λ/2πε₀r
where;
- λ is linear charge density
- ε₀ is permitivity of free space
- r is the position of the charge
<h3>At a distance of 5.5 m</h3>

<h3>At a distance of 2.5 m</h3>

Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.
Learn more about electric field here: brainly.com/question/14372859
Answer:

Explanation:
given,
In first case Volume remains constant.
Work done in the first case is zero.
In Second case Volume change
V₁ = 0.2 m³
V₂ = 0.11 m³
Pressure, P = 5.5 x 10⁵ Pa
Work done = Pressure x change in volume
W = P ΔV



Hence, Work done when volume changes is equal to 
I think the answer is <span>AM radio </span><span>waves. Im not sure but if its right can you mark my brainliest?
</span>
The focal length of the lens is 25cm
given:
power,p=4 diopters
what is focal length?
Focal length is the distance between the point of convergence of your lens and the sensor or film recording the image.
what is diopter?
The unit of optical power of lens is called diopter.It is the optical power of the lens.
we know,
p=1/f
where,
p= power
f= focal length
f=1/p
f=1/4
=0.25m
=25cm
Thus,the focal length of the lens is 25cm
learn more about focal length from here: https/brainly.com/question/28203589
#SPJ4