Questions
Two similar point charges are placed at a distance r apart in air. Assume that a slab of thickness one third the separation between the charges is placed between the charges and it is observed that the ratio of Coulomb’s repulsive force before and after placement of dielectric is 25:9. Then the dielectric constant K of such a slab is
detailed solution
Correct option is C
F1=q1q2d214πεoIn the second case effective distance2d3+kd3F2=14πEoq1q22d3+kd32F1F2=2.59=q1q2/d2q1q2/2d3+kd322.59=2d3+kd32d2=53=23+k133=kK=9Similar Questions
In the circuit shown, the base current is 30 µA. The value of is (Neglect )
Get your all doubts cleared from
Sri Chaitanya experts
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests