Consider a sphere of radius R which carries a uniform charge density ρ . If a sphere of radius R 2 is carved out of it, as shown, the ratio E A → E B → of magnitude of electric field E A → and E B → , respectively, at points A and B due to the remaining portion is:

Question

Consider a sphere of radius R which carries a uniform charge density &#961; . If a sphere of radius R 2 is carved out of it, as shown, the ratio E A &#8594; E B &#8594; of magnitude of electric field E A &#8594; and E B &#8594; , respectively, at points A and B due to the remaining portion is:

Aicharya · Accepted Answer

E = &#961; r 3 &#949; 0 E A = - &#961; R 2 3 &#949; 0 E A = &#961; R 6 &#949; 0 Electric filed at point B = E B = E 1 A + E 2 A E 1 A = Electric Filed Due to solid sphere of radius R at point B = &#961; R 3 &#949; 0 E 2 A = Electric Filed Due to solid sphere of radius R 2 (which having charge density &#8211; &#961; ) at point B = - K Q ' &#215; 4 9 R 2 = - &#961; R 54 &#949; 0 E B = E 1 A + E 2 A = &#961; R 3 &#949; 0 - &#961; R 54 &#949; 0 = 17 &#961; R 54 &#949; 0 E A E B = 9 17

Consider a sphere of radius R which carries a uniform charge density ρ . If a sphere of radius R 2 is…

Solution

Consider a sphere of radius R which carries a uniform charge density &#961; . If a sphere of radius R 2 is…

Solution

Consider a sphere of radius R which carries a uniform charge density ρ . If a sphere of radius R 2 is…