\[ a )
\frac { \sigma (R + r) } { \epsilon _{ o } } \]
\[ b )
\frac { \sigma (R - r) } { \epsilon _{ o } } \]
\[ c )
\frac { \sigma (R - r) } { 2 \epsilon _{ o } } \]
\[ d )
\frac { \sigma (R - r) } { 4 \epsilon _{ o } } \]
Solution:
Electric potential ( Definition ) : The electric potential at a
point in electric field is defined as the amount of work done to bring unit
positive charge from infinite distance to that point against the direction of
electric intensity.
I . e .
\[ V = \frac { w } { q_{ o } } \]
Where \[q _{ o } \]is a unit charge
i.e \[q _{ o } =1\]
- - - - - - - - - (
1 )
But ,
\[W = \frac { q } { 4 \pi \epsilon _{o} r }\]
\[V = \frac { q } { 4 \pi \epsilon _{o} r }\]
- - - - - - - - - ( 2 )
The potential at center due to sphere 1 is
\[V _ { 1 } = \frac { q } { 4 \pi \epsilon _{ o } R }\]
-
- - - - - - - - ( 3 )
And due to sphere 2 is:
\[V _ { 2 } = \frac { q } { 4 \pi \epsilon _{ o } r }\]
-
- - - - - - - - ( 4 )
Therefore the electric potential at common centre can be written
as
\[V = V_{1}+ V_{2}\]
\[V = \frac { q } { 4 \pi
\epsilon _{ o } R } + \frac { q } { 4 \pi \epsilon _{ o } r }\]
\[ V =\frac { q } { 4 \pi \epsilon _{ o } } ( \frac { 1 } { R } +
\frac { 1 } { r } ) \]
\[ V =\frac { q } { 4 \pi \epsilon _{ o } } ( \frac { R+r } { Rr
} ) \]
- - - - - - - - - ( 5 )
But σ is surface charge density i . e .
\[\sigma = \frac{ ( total charge on the surface of sphere
q ) } {( surface area of sphere ) } \]
Therefore ,
Equation ( 5 ) becomes
\[V=\sigma \frac { ( R + r ) } { \epsilon _{ O } } \]
The correct option is : option [ a ]
Hi. I’m Designer of Blog Magic. I’m CEO/Founder of ThemeXpose. I’m Creative Art Director, Web Designer, UI/UX Designer, Interaction Designer, Industrial Designer, Web Developer, Business Enthusiast, StartUp Enthusiast, Speaker, Writer and Photographer. Inspired to make things looks better.
0 comments:
Post a Comment