- Home
- Standard 12
- Physics
1. Electric Charges and Fields
medium
A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of $volt-meter$ associated with the curved surface $B,$ the flux linked with the plane surface $A$ in units of $V-m$ will be
A$\;\frac{q}{{2{\varepsilon _0}}}$
B$\frac{\phi}{3}$
C$\;\frac{q}{{{\varepsilon _0}}}-\phi$
D$\frac{1}{2}\left(\frac{ q }{\varepsilon_0}-\phi\right)$
(AIPMT-2007) (AIIMS-2008)
Solution
Let ${\phi _A},{\phi _B}$ and ${\phi _C}$ are the electric flux linked with $A,B$ and $C.$
According to gauss theorem,
${\phi _A} + {\phi _B} + {\phi _C} = \frac{q}{{{\varepsilon _0}}}$
$\sin ce\,{\phi _A} = {\phi _C},$
$\therefore \,2{\phi _A} + {\phi _B} = \frac{q}{{{\varepsilon _0}}}\,\,\,or\,\,2{\phi _A} = \frac{q}{{{\varepsilon _0}}} – {\phi _B}$
or, $2{\phi _A} = \frac{q}{{{\varepsilon _0}}} – \phi $
(Given ${\phi _B} = \phi $).
$\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} – \phi } \right).$
According to gauss theorem,
${\phi _A} + {\phi _B} + {\phi _C} = \frac{q}{{{\varepsilon _0}}}$
$\sin ce\,{\phi _A} = {\phi _C},$
$\therefore \,2{\phi _A} + {\phi _B} = \frac{q}{{{\varepsilon _0}}}\,\,\,or\,\,2{\phi _A} = \frac{q}{{{\varepsilon _0}}} – {\phi _B}$
or, $2{\phi _A} = \frac{q}{{{\varepsilon _0}}} – \phi $
(Given ${\phi _B} = \phi $).
$\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} – \phi } \right).$
Standard 12
Physics
Similar Questions
medium
