The electric field in a region is given by E → = 3 5 E 0 i ^ + 4 5 E 0 j ^   N   C - 1 . The ratio of flux of reported field through the rectangular surface of area 0 . 2   m 2 (parallel to y - z plane) to that of the surface of area 0 . 3   m 2 (parallel to x - z plane) is a : b = a : 2 , where a = ? [Here i ^ ,   j ^ and k ^ are unit vectors along x ,   y and z -axes respectively

Step-by-step solution

E → = 3 E 0 5 i ^ + 4 E 0 5 j ^   N   C - 1 A 1 = 0 . 2   m 2 [parallel to y - z plane] = A → 1 = 0 . 2   m 2 i ^ A 2 = 0 . 3   m 2 [parallel to x - z plane] A → 2 = 0 . 3   m 2 j ^ Now, ϕ a = 3 E 0 5 i ^ + 4 E 0 5 j ^ · 0 . 2 i ^ = 3 × 0 . 2 5 E 0 &   ϕ b = 3 E 0 5 i ^ + 4 E 0 5 j ^ · 0 . 3 j ^ = 4 × 0 . 3 5 E 0 Now, ϕ a ϕ b = 0 . 6 1 . 2 = 1 2 = a b ⇒ a : b = 1 : 2 ⇒ a = 1