The electric potential V(z) on the z-axis is : V = 
The magnitude of the electric field on the z axis is : E = kб 2
( 1 - [z / √(z² + a² ) ] )
<u>Given data :</u>
V(z) =2kQ / a²(v(a² + z²) ) -z
<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>
Considering a disk with radius R
Charge = dq
Also the distance from the edge to the point on the z-axis = √ [R² + z²].
The surface charge density of the disk ( б ) = dq / dA
Small element charge dq = б( 2πR ) dr
dV 
----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv = 
V = ![\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]](https://tex.z-dn.net/?f=%5Cpi%20k%5Calpha%20%5B%20%28a%5E2%2Bz%5E2%29%5E%5Cfrac%7B1%7D%7B2%7D%20-z%20%5D)
= ![\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]](https://tex.z-dn.net/?f=%5Cpi%20k%20%28%5Cfrac%7BQ%7D%7B%5Cpi%20%5Calpha%20%5E2%7D%29%5B%28a%5E2%20%2Bz%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%20-z%20%5D)
Therefore the electric potential V(z) =
Also
The magnitude of the electric field on the z axis is : E = kб 2( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
Learn more about electric potential : brainly.com/question/25923373
In which it is the plane in which charges move. for example in electromagnetic waves polarization<span> means the plane in which electrons oscillate. ... it is the distribution of electric charges (electron) in an object. </span>induction<span> of charges takes place in some insulators just by some mechanical interaction such as rubbing</span>
Answer:
1.34 x 10^3 Pa
Explanation:
density of oil = 0.85 x 10^3 kg/m^3
g = 9.81 m/s^2
height of oil column = 16.1 cm = 0.161 m
Pressure on the surface of water = height of oil column x density of oil x g
= 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa
Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.
<h2>
Answer:</h2>
0.126m
<h2>
Explanation:</h2>
According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;
F = k x e -------------------(i)
Where;
k = the spring's constant.
From the question, the force acting on the spring is the weight(W) of the mass. i.e
F = W -----------------------(ii)
<em>But;</em>
W = m x g;
where;
m = mass of the object
g = acceleration due to gravity [usually taken as 10m/s²]
<em>From equation (ii), it implies that;</em>
F = W = m x g
<em>Now substitute F = m x g into equation(i) as follows;</em>
F = k x e
m x g = k x e ------------------(iii)
<em>From the question;</em>
m = m1 = 3.5kg
k = 278N/m
<em>Substitute these values into equation (iii) as follows;</em>
3.5 x 10 = 278 x e
35 = 278e
<em>Now solve for e;</em>
e = 35/278
e = 0.126m
Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m
True
In fact, the weight of an object on the surface of the Earth is given by:
where m is the mass of the object and
is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
