Two concentric spheres kept in air have radii R and r. They have similar charge and equal surface charge density `sigma`. The electrical potential at their common centre is (where, `epsi_(0) =` permittivity of free space)A. `(sigma(R+r))/(epsilon_(0))`B. `(sigma(R-r))/(epsilon_(0))`C. `(sigma(R+r))/(2epsilon_(0))`D. `(sigma (R+r))/(4epsilon_(0))`

Two concentric spheres kept in air have radii R and r. They have simil

1.	Two concentric spheres kept in air have radii R and r. They have similar charge and equal surface charge density `sigma`. The electrical potential at their common centre is (where, `epsi_(0) =` permittivity of free space)A. `(sigma(R+r))/(epsilon_(0))`B. `(sigma(R-r))/(epsilon_(0))`C. `(sigma(R+r))/(2epsilon_(0))`D. `(sigma (R+r))/(4epsilon_(0))`
Answer» Correct Answer - A Potential at the centre due to sphere 1, `V_(1)=(1)/(4pi epsilon_(0))(q)/(R)` and due to sphere 2 is `V_(2)=(1)/(4pi epsilon_(0))(q)/(r)` `therefore` Electric potential at the common centre `V=V_(1)+V_(2)=(1)/(4pi epsilon_(0))(q)/(R)+(1)/(4pi epsilon_(0))(q)/(r)` `V=(q)/(4pi epsilon_(0))[(1)/(R)+(1)/(r)]=(q)/(4pi epsilon_(0))[(R+r)/(Rr)]` we have `(q)/(4pi Rr)=sigma` `rArr" Electric potential, V"=(sigma(R+r))/(epsilon_(0))`

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