a solid conducting sphere of radius …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Test
Related Questions
Posted by Khushbu Otti 1 year, 7 months ago
- 0 answers
Posted by Anterpreet Kaur 1 year, 8 months ago
- 0 answers
Posted by Aniket Mahajan 1 year ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 6 years, 8 months ago
The whole charge is transferred to the outer sphere.
Therefore, Heat generated = Ui - Uf
{tex}H = \frac { q ^ { 2 } } { 2 C _ { 1 } } - \frac { q ^ { 2 } } { 2 C _ { 2 } }{/tex}
Here, C1 = {tex}4 \pi \varepsilon _ { 0 } R _ { 1 }{/tex}
and C2 = {tex}4 \pi \varepsilon _ { 0 } R _ { 2}{/tex}
{tex}\therefore H = \frac { q ^ { 2 } } { 8 \pi \varepsilon _ { 0 } } \left( \frac { R _ { 2 } - R _ { 1 } } { R _ { 1 } R _ { 2 } } \right){/tex}
{tex}= \frac { \left( 20 \times 10 ^ { - 6 } \right) ^ { 2 } \left( 9 \times 10 ^ { 9 } \right) ( 0.1 ) } { ( 2 ) ( 0.1 ) ( 0.2 ) } = 9 J{/tex}
1Thank You