Explanation:
The given data is as follows.
q = 6.0 nC = 
inner radius (r) = 1.0 cm = 0.01 m (as 1 cm = 100 m)
So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.
Formula to calculate the charge density is as follows.
.......... (1)
Since, area of the sphere is as follows.
A = 
........... (2)
Hence, substituting equation (2) in equation (1) as follows.
= 
= 
or, = 4.77 
Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 .
This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:
D = 1/2 A T²
Distance = (1/2) (acceleration) (time²)
The reason I never forgot it is because it's SO useful SO often. You really should memorize it. And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)
The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth. Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .
In 5 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (5 sec)²
D = (4.9 m/s²) (25 sec²)
D = 122.5 meters
In 6 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (6 sec)²
D = (4.9 m/s²) (36 sec²)
D = 176 meters
Answer:
(a) 8 V, (b) 144000 V, (c) 2 x 10^(-8) C
Explanation:
(a) charge, q = 5 μC , Work, W = 40 x 10-^(-6) J
The electric potential is given by
W = q V
(b)
charge, q = 8 x 10^(-6) C, distance, r = 50 cm = 0.5 m
Let the potential is V.
(c)
Work, W = 8 x 10^(-5) J, Potential difference, V = 4000 V
Let the charge is q.
W= q V
<h2>HOPE THIS HELPS YOU ....</h2><h3>PLEASE MARK ME AS BRAINILIST...</h3>
Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given 
The surface area to volume ratio for right circular cylinder is given 
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
