Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respectively. Determine the electric field for (a) 0 < r < r1 , (b) r1 < r < r2 and (c) r > r2. (d) Under what conditions

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4  =

So,

Rearranging the above equation to get Electric field, we will get:

E =

Multiply and divide by

E = x

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x ) /( x )

c) r > r2 :

Electric Field = ?

E x 4  =

Rearranging the above equation for E:

E =

E = +

As we know from above, that:

=  (σ1 x ) /( x )

Then, Similarly,

= (σ2 x ) /( x )

So,

E = +

Replacing the above equations to get E:

E = (σ1 x ) /( x ) + (σ2 x ) /( x )

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x = - σ2 x

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s