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Potential difference beween centre and surface of the sphere of radius R and uniorm volume charge density `rho` within it will beA. `(rhoR^2)/(6epsilon_0)`B. `(rhoR^2)/(4epsilon_0)`C. `(rhoR^2)/(3epsilon_0)`D. `(rhoR^2)/(2epsilon_0)` |
Answer» Correct Answer - A `rho=a/((4//3)piR^3)` `:.q=4/3pirhoR^3` `V_C+V_S=3/2(1/(4piepsilon_0).q/R)-1/(4piepsilon_0).q/R` `=q/(8piepsilon_0R)` Substituting the value of `q` we have `V_C-V_S=(rhoR^3)/(6epsilon_0)` |
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