Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density $σ$ . They are brought in contact and separated. What will be the new surface charge densities on them?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$σ_{1} = \frac{5}{2} σ, σ_{2} = \frac{5}{6} σ$

$σ_{1} = \frac{5}{6} σ, σ_{2} = \frac{5}{2} σ$

$σ_{1} = \frac{5}{3} σ, σ_{2} = \frac{5}{6} σ$

$σ_{1} = \frac{5}{2} σ, σ_{2} = \frac{5}{3} σ$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$Q_{1} = σ 4 {πR}_{1}^{2} = σ 4 {πR}^{2}; Q_{2} = σ 4 π (2 R)^{2} = σ 16 {πR}^{2};$

$After Redistribution of charges \frac{Q_{1}^{'}}{Q_{2}^{'}} = \frac{R}{2 R} \Rightarrow Q_{2}^{'} = 2 Q_{1}^{'} . . (i); a n d, Q_{1}^{'} + Q_{2}^{'} = 20 {σπR}^{2} . . . (ii)$

$From eq . (i) and (ii) \begin{array}{l} Q_{1}^{'} = \frac{20}{3} {σπR}^{2} \Rightarrow σ_{1}^{'} = \frac{5}{3} σ; \\ Q_{2}^{'} = \frac{40}{3} {σπR}^{2} \Rightarrow σ_{2}^{'} = \frac{5}{6} σ; \end{array}$