What mediums are best for light waves?
1 answer:
You might be interested in
Answer:
(a) E=0 : 0 cm from the center of the sphere
(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere
(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere
(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere
Explanation:
If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere
: Formula (1) To calculate the electric field in the region outside the sphere r ≥ a
:Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a
Where:
K: coulomb constant
a: sphere radius
Q: Total sphere charge
r : Distance from the center of the sphere to the region where the electric field is calculated
Equivalences
1μC=10⁻⁶C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
Q=16.2 μC=16.2 *10⁻⁶C
a= 40 cm = 40*10⁻²m = 0.4m
Problem development
(a)Magnitude of the electric field at 0 cm :
We replace r=0 in the formula (2) , then, E=0
(b) Magnitude of the electric field at 10.0 cm from the center of the sphere
r<a , We apply the Formula (2):
E= 227.8*10³ N/C
(c) Magnitude of the electric field at 40.0 cm from the center of the sphere
r=a, We apply the Formula (1) :
E= 911.25*10³ N/C
(d) Magnitude of the electric field at 59.5 cm from the center of the sphere
r>a , We apply the Formula (1) :
E= 411.84 * 10³ N/C