Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m).
The point then lies on the y-axes at d = 0.03 m.
from symmetry, the field at that point will be ascending along the y-axes.
A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point.
Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength.
All in all, the infinitesimal field strength from the charge between x and x+dx is:
dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2)
Therefore, upon integration,
E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2.
This gives:
E = k lambda L / (d sqrt((L/2)^2 + d^2) )
But lambda L = Q, the total charge on the rod, so
E = k Q / ( d * sqrt((L/2)^2 + d^2) )
A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.
Answer:
The intensity increased by a factor of 158489
Explanation:
Given that,
Sound level = 95.0 dB
Sound level = 43.0 dB
Frequency = 10000 Hz
We need to calculate the ratio of sound intensity
Using formula of sound level

Put the value into the formula
...(I)
.....(II)
Subtracting these equations


Taking inverse log

Hence, The intensity increased by a factor of 158489
Answer:

is the no. of electrons
Explanation:
Given:
- quantity of charge transferred,

<u>No. of electrons in the given amount of charge:</u>
As we have charge on one electron 
so,


is the no. of electrons
- Now if each water molecules donates one electron:
Then we require
molecules.
<u>Now the no. of moles in this many molecules:</u>

where
Avogadro No.


- We have molecular mass of water as M=18 g/mol.
<u>So, the mass of water in the obtained moles:</u>

where:
m = mass in gram

