Answer:
Explanation:
We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .
∫ E ds = q / ε
Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so
∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .
This also represents total flux coming out of the charge q on all sides .
This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .
If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .
It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors
1 ) charge q and
2 ) the permittivity of medium ε around .
Answer:
material work function is 0.956 eV
Explanation:
given data
red wavelength 651 nm
green wavelength 521 nm
photo electrons = 1.50 × maximum kinetic energy
to find out
material work function
solution
we know by Einstein photo electric equation that is
for red light
h ( c / λr ) = Ф + kinetic energy
for green light
h ( c / λg ) = Ф + 1.50 × kinetic energy
now from both equation put kinetic energy from red to green
h ( c / λg ) = Ф + 1.50 × (h ( c / λr ) - Ф)
Ф =( hc / 0.50) × ( 1.50/ λr - 1/ λg)
put all value
Ф =( 6.63 ×
(3 ×
) / 0.50) × ( 1.50/ λr - 1/ λg)
Ф =( 6.63 × (3 × ) / 0.50 ) × ( 1.50/ 651×
- 1/ 521 ×)
Ф = 1.5305 ×
J × ( 1ev / 1.6 × J )
Ф = 0.956 eV
material work function is 0.956 eV
Ball thrown into the air at an angle.
C., used in power plants I think.
