Electric Current
I need the answer to question no.8 .It is previous years jee advanced question.
1 Answer
Saurav
BY GAUSS LAW
the surface electric field on the surface of outermost sphere will be
∫E.ds=(Q1+Q2+Q3)/
0
THUS TOTAL CHARGE ON THE SURFACE WILL BE Q1+Q2+Q3
SIMILARLY THE CHARGE ON MIDDLE AND INNER SPHERES WILL BE Q1+Q2,Q1 RESPECTIVELY
BUT ALL OF THEM HAS GOT THE SAME CHARGE DENSITY
THEREFOR EQUATE THERE SURFACE CHARGE DENSITIES
YOU WILLL GET (Q1+Q2+Q3)/9 =(Q1+Q2 )/4 =Q1
THEN YOU WILL GET THE RATIO 1:3:5
PLEASE APPROVE
Other Related Questions on electric current
The fig. Shows a network of resistance in which the point A is earthed
The fig. Shows a network of resistance in which the point A is earthed
electric current
Last Activity: 4 Years ago
When a p.d of 2v is applied across a resistor, 10j of energy are covered. What charge flows through the resistor?
When a p.d of 2v is applied across a resistor, 10j of energy are covered. What charge flows through the resistor?
electric current
Last Activity: 4 Years ago
Please provide it's solution. In video form Find the equivalent resistance of the given figure
Please provide it's solution. In video form
Find the equivalent resistance of the given figure
electric current
Last Activity: 4 Years ago
The resistivity of material is 10 ohm m at 30 C when the temperature drops to 28 then the resistivity drops by 2. The temperature coefficient of resistivity of material is
The resistivity of material is 10 ohm m at 30 C when the temperature drops to 28 then the resistivity drops by 2. The temperature coefficient of resistivity of material is
electric current
Last Activity: 4 Years ago
Three cells of 2.0v each and internal resistance of 1.0 each are connected in series and in parallel calculate the new emf and new internal resistance in each connection
Three cells of 2.0v each and internal resistance of 1.0 each are connected in series and in parallel calculate the new emf and new internal resistance in each connection
electric current
Last Activity: 4 Years ago