Answer:
when the radius is halved, F becomes 4 times
Explanation:
<u><em>The formula is given by</em></u>

<u><em>When r is halved:</em></u>

=> 
=> 
<u><em>This means when the radius is halved, F becomes 4 times</em></u>
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
Answer:
c
Explanation:
though c is wider it has more water.
Answer:
No. 67
Peter Street
12th Road
Chennai
24th June 201_
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Yours affectionately
your name