Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
A ball of radius R carries a positive charge whose volume density at a point is given by Where ρ 0 is a constant and r is distance from the centre. The value of maximum electric field due to this configuration is A ball of radius R carries a positive charge whose volume density at a point is given by Where ρ0 is a constant and r is distance from the centre. The value of maximum electric field due to this configuration is
A ball of radius R carries a positive charge whose volume density at a point is given by Where ρ0 is a constant and r is distance from the centre. The value of maximum electric field due to this configuration is
(a) We assume the ball to be divided into infinite number of concentric thin shell of thickness dr. Let us assume such a shell at a radial distance r. Volume of this shell =pdV=4πr 2ρdr=4πr2 ρ0 (1− Rr)dr Net charge enclosed in the sphere of radius r q=∫pdv= 0∫r? 4πρ02 (1− Rr?)dr=4πρ 0? ( 3r 3? − 4Rr 4? ) Electric field at radial distance r E= 4πε0 r 2q? = 4πε 0 r 21 ×4πρ 0?( 3r 3? − 4Rr 4 )= 3ε 0? ρ 0? r (1− 4R3r) Now for those points outside the sphere, we have to take into account the total charge contained in the sphere of radius R. Total charge Q= 0∫R ρ 0? (1− Rr? )4πr 2 dr= 3πρ 0? R3 Electric field, using Gauss's laws, at a point which is at distance r(>R) from the centre; ∫E.ds= ε 0 Q? ?E= 4πε 0 r 2Q? = 2ε 0 r 2ρ 0? R3 (b) Again let us consider the expression for electric field within the sphere of radius R: E= 3ε 0ρ 0 r? (1− 4R3r) For maximum electric field: drdE? =0?1− 4R6r =0?r= 32R ?Putting, r= 3 2R , value of maximum electric field, E max? = 9∈ 0?ρ 0? R
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -