TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Average intensity, Iavg=‌

Pavg

A

where, P‌avg ‌= average power generated by source
=4%‌ of ‌100W=‌

4

100

×100W=4W
A= area to which energy is transmitted
=4πr2=4π×(22=4π×4=16πm2
On putting the values into Eq. (i), we have
Iavg=‌

4

16π

⇒Iavg=‌

1

4π

Also average intensity, I‌avg ‌=ε0Erms2c
where, ε0= permittivity of free space
and Erms= root mean square (rms) value of electric field.
c= speed of light in vacuum =3×108m∕s
On putting the values into Eq. (iii), we get
I‌avg ‌=ε0×Ems2×3×108
On equating both the Eqs. (ii) and (iv), we get

1

4π

=ε0×Erms2×3×108⇒‌

1

4πε0×3×108

=Erms2
Erms2=‌

1

4πε0

×‌

1

3×108

=9×109×‌

1

3×108

(∵‌

1

4πε0

=9×109Nm2∕C2)
=3×10=√30V∕m