The power crossing a surface perpendicular is equal to theintensity at the surfa

Question

The power crossing a surface perpendicular is equal to theintensity at the surface times the area of the surface. Since thesource emits uniformly in all directions, the light intensity isthe same at all points on the imaginary spherical surface.Moreover, the light crosses this surface perpendicularly. S =c0E2 relates the average light intensity at the surface to its rmselectric field strength (which is known), and the area of thesurface can be found from a knowledge of its radius.

A light bulb emits light uniformly in all directions. The averageemitted power is 150.0 W. At a distance of 5.30 m from the bulb, determine (a) the averageintensity of the light, (b) the rms value of the electric field,and (c) the peak value of the electric field. (a) S = Enter anumber. 12---Select---JWW/m^2J/m3 (b) Erms = Enter anumber. 34---Select---TN/CJNW (c) E0 = Enter anumber. 56---Select---NTN/CWJ S =c0E2 (a) S = Enter anumber. 12---Select---JWW/m^2J/m3 (b) Erms = Enter anumber. 34---Select---TN/CJNW (c) E0 = Enter anumber. 56---Select---NTN/CWJ Enter anumber. Enter anumber. Enter anumber. (a) S = Enter anumber. 12---Select---JWW/m^2J/m3 (b) Erms = Enter anumber. 34---Select---TN/CJNW (c) E0 = Enter anumber. 56---Select---NTN/CWJ

Explanation / Answer

Average intensity of the light. P=150/(45,3^2)=0,425(W/m2). --------- P=0*E^2. so E=2,19e5(V/m) (RMS value). -------- E0=2,19e5*sqrt(2)=3,1e5(V/m) (peak value)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.