SME
Electromagnetic Theory · Problem type
A point source: intensity and field at a distance
Use this for “A 5 W radio transmitter sends its power evenly in all directions. Find the intensity and the electric field amplitude 300 m away.”
Table
Distance (r)
(m)
Electric field amplitude (E₀)
(V/m)
This is an example. Type your own values to start a new problem.
I
=
P (
)
4π
(
r (
)
)
2
=
I (W/m²)
I
=
½cε₀
(
E₀ (V/m)
)
2
Formulas
I
=
P
4π ×
r
2
1.989
=
100
4π ×
2
2
I
=
1
2
×
3.00 ×
10
8
×
8.85 ×
10
-12
×
E₀
2
1.989
=
1
2
×
3.00 ×
10
8
×
8.85 ×
10
-12
×
38.71
2
Assumptions
•
The source is small next to r and sends its power evenly in every direction; nothing absorbs or reflects it on the way.
•
All of P crosses a sphere of area 4πr², so I = P ÷ (4πr²): twice as far, a quarter of the intensity.
•
Far from the source the wave is nearly plane, so I = ½cε₀E₀² with c = 3.00 × 10⁸ m/s and ε₀ = 8.85 × 10⁻¹² F/m.
Step-by-step
WE KNOW
P = 100 W, r = 2 m
FIND
intensity (I), electric field amplitude (E₀)
Step 1 · Find intensity (I)
Use I = P ÷ (4π × r²)
The
power
spreads
over
a
sphere
of
radius
r
:
divide
it
by
the
sphere’s
area
4πr².
I
=
100
4π ×
2
2
I
=
1.989
W/
m
2
Step 2 · Find electric field amplitude (E₀)
Use I = ½ × 3.00 × 10⁸ × 8.85 × 10⁻¹² × E₀²
The
intensity
is
1
2
cε₀
E₀
2
,
the
Poynting
vector
averaged
over
a
cycle:
solve
it
for
E₀
.
E₀
=
√(2
×
I
3.00 ×
10
8
× 8.85 ×
10
-12
)
E₀
=
√(2
×
1.989
3.00 ×
10
8
× 8.85 ×
10
-12
)
E₀
=
√(2
×
1.989
0.002655
)
E₀
=
√(
3.978
0.002655
)
E₀
=
√
1
,498.3
E₀
=
38.71
V/m
Check
Put the numbers back in. Both sides should match.
1.989
=
100
4π ×
2
2
✓
1.989
=
1
2
×
3.00 ×
10
8
×
8.85 ×
10
-12
×
38.71
2
✓
Topic check
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