A thin insulating rod is hinged about one of its ends. It can rotate on a smooth horizonal surface. The charge density on the rod is defined as =15 x ^2, 0 x 2 =- bx ^ n , 2 x where b is a positive constant. An electric field E ₀ in the horizontal direction and perpendicular to the rod is switched on. Find the value of (b+n)^2 , if the rod has to remain stationary.

Step-by-step solution

For 0 x 2 , ~d ₁= E ₀( dx ) x ₁= ₀^ / 2 E ₀ (15 x ^2 dx ) x =15 E ₀ ₀^ / 2 x ^3 dx = 15 E ₀ 4 ( 4 )^4= 15 E ₀ ^4 64 For 1 2 x , d ₂=E₀ (b x^n d x ) x ₂= _ / / 2 E₀ b x^ n+1 d x= E₀ b n+2 (2^ n+2 -1 ) 2^ n+2 ^ n+2 Now According to question ₁= ₂ 15 E ₀ ^4 64 = E ₀ ~b (2^ n +2 -1 ) ^ n+2 ( n +2) 2^ n +2 4= n +2 n =2 and b =1 Therefore ( a + b )^2=(2+1)^2=9