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This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
Using the Newton's Secound Law, we have:
Number 3
If you notice any mistake in my english, please let me know, because i am not native.
Using the Newton's Secound Law, we have:
Number 3
If you notice any mistake in my english, please let me know, because i am not native.
Answer:
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