Answer:
66.375 x 10⁻⁶ C/m
Explanation:
Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as
;
∅ = Q / ε₀ -----------------(i)
Where;
∅ = 7.5 x 10⁵ Nm²/C
ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²
Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;
7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)
Solve for Q;
Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²
Q = 66.375 x 10⁻⁷ C
Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e
L = Q / l ----------------------(ii)
Where;
Q = 66.375 x 10⁻⁷ C
l = length of the rod = 10.0cm = 0.1m
Substitute these values into equation (ii) as follows;
L = 66.375 x 10⁻⁷C / 0.1m
L = 66.375 x 10⁻⁶ C/m
Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.
Answer:
Explanation:
The angular position of first intensity maxima is given by the expression
= λ / d where λ is wave length of light and d is slit separation
putting the values given in the problem. ,
.02 = λ₁ / D
Now wavelength has been increased to 2λ₁
angular separation of first maxima
= 2λ₁ /D
= 2 x .02
= .04 radians.
