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Two sphere of radius R and 2R are charged, so that both of these have same surface charge density σ. The spheres are located away from each other and are connected by a thin conducting wire. Find the new charged density on sphere of radius R.

riya singhania , 8 Years ago
Grade 12
anser 5 Answers
Pushkar Prateek

Last Activity: 8 Years ago

Let Q1 and Q2 be the initial charges on the spheres.Since charge density is same
Q1/4\piR2  = Q2/4\pi(2R)2
So, Q1/Q2 = 1:4
Or we can say the total charge of the system is 5Q1.
When we join any two conductors with a conducting wire their potential will become same.
Let final charges on them be X1  and X2 respectively.
K X1 / R   = K X2 / 2R
 X1 / X2 = 1:2
Or we can say the total charge of the system is 3X1.
From conservation of charge of system
 5Q1 = 3X1  
Initial charge density Q1/4\piR2 =\sigma
Hence final charge density X1/4\piR2 =  5\sigma/3.(Calculation mistake above.)

Shivangi Baral

Last Activity: 6 Years ago

Let q1, q2 be initial charges on spheres with radius R and 2R respectively.
As charge surface charge density of both the spheres is same 
So sigma1=sigma
=> q1/4pir^2 = q2/4pi(2r)^2
=> 4q1=q2
=>total charge = q1+q2 = 5q1
 
when they are joined by a conducting wire , they will have same potential on their surfaces ,
let charges now on each sphere be Q1 and Q2 respectively
=> kQ1/R = kQ2/2R
=> 2Q1=Q2
total charge = Q1+ Q2 = 3Q2
 
as total charge of the system remains conserved
=> 5q1=3Q2 = 3(2Q1) =6Q1
=> Q1 = 5q1/6
 
hence final charge density = kQ1/R = 5kq1/6R
 

Shivangi Baral

Last Activity: 6 Years ago

in above answer its Q1 + Q2 = 3Q1
=> 5q1=3Q1
=> Q1=5q1/3 
=>kQ1/R = 5Kq1/6R
sorry for the above mistake :)

Ajeet Tiwari

Last Activity: 4 Years ago

hello student
642-865_upload 42.png

Hope it helps
Thankyou

Yash Chourasiya

Last Activity: 4 Years ago

Dear Student

Please see the solution in the attachment.
643-2416_Untitled.png

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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