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@metal
I agree with u. This is the correct explanation.

AND FIRST OF ALL CORRECT UR QUESTION WHERE U HAVE WRITTEN "DO ELECTRONS ACTUALLY FLOW FROM HIGHER POTENTIAL TO LOWER POTENTIAL................".
ACTUALLY ELECTRONS ALWAYS FLOW FROM LOWER POTENTIAL TO HIGHER POTENTIAL AND IT IS POSITIVE CHARGE OR WE CAN SAY CURRENT THT ACTUALLY FLOWS FROM HIGHER POTENTIAL TO LOWER POTENTIAL.

SEE,
FOR THIS SUPPOSE TWO CONDUCTORS , ONE AT HIGHER POTENTIAL AND OTHER AT LOWER POTENTIAN ARE CONNECTED BY A WIRE. NOW THE FLOW OF ELECTRONS WILL BE ONLY DUE TO DIFFERENCE IN POTENTIAL NOT DUE TO DIFFERENCE IN ENERGIES. AND IT GOES LIKE "THE FLOW OF CHARGE WILL TAKE PLACE UNTILL BOTH THE CONDUCTORS BECOME EQUIPOTENTIAL. SO, TO EQUALISE THE POTENTIAL THE ELECTRONS MUST ACTUALLY FLOW FROM ONE CONDUCTOR TO ANOTHER THT IS FROM LOWER POTENTIAL TO HIGHER POTENTIAL.

RATE IF SATISFIED OR U CAN ASK FURTHER.

@reddevil
hey dont u think u must have given charge on the shell because u have asked the deflection only due to effect of electric field. there will be deflection ofcourse but only due to gravity. Since mass of the box is given the effect of gravity cant be ignored and the deflection will be only due to tht and in tht case it will be a question of projectile motion unless the shell have some charge if we are can neglect the induction!!!!!!!!!!!!!!!!!

nevertheless plz.. let me know if the question goes same like this and u want to ask the same!!!!

the forces acting on the hanging plate will be-

1. tension T upwards

2. weight mg downwards

3. elctric force qE downwards

now

now for the equilibrium of the plate

0r,

therefore,

THIS WILL BE THE REQUIRED VALUE OF m

HOPE THIS IS THE CORRECT ANSWER.

LET ME KNOW IF THIS IS CORRECT OR SOME ONE RECTIFY ME

PLZ.. VOTE IF I M CORRECT.

AS FAR AS PHYSICS IS CONCERNED IN MY VIEWS THE PRIORITY LIST COULD BE-
1. MODERN PHYSICS
2. MAGNETIC EFFECT OF CURRENT
3. FLUIDS
4. DYANAMICS
5. CURRENT ELECCTRICITY

THESE ARE THE TOPICS WHICH IF U MASTER CAN MAKE U DO THE PHYSICS PORTION WELL TO GET SELECTED.
BUT FOR GOOD RANK COMPLETE COVERAGE OF THE SYLLABUS IS A MUST.

OKKK NOW.
SEE, NOW WHEN WE HAVE GOT THE ORIGINAL QUESTION IT CAN BE VERY CLEAR.
THERE WOULD BE NO CAPACITOR FORMED BY THE PLATES 2 AND 3 COZ. THEY ARE CONNECTED AND HENCE THEY SHOULD BE CONSIDERED AS A SINGLE CONDUCTOR AND NOT TWO DIFFERENT PLATES . THEREFORE THJEY DEFINITELY DO NOT FORM A CAPACITOR.

@harsh
OK HARSH LETS COUNTER UR ARGUMENT. SEE, LET US TAKE AN EXAMPLE OF A ONE LANE ROAD. NOW IN THE FRONT THERE IS A BYCYCLE, AFTER IT A BIKE AND THEN BEHIND IT A CAR. NOW THE CAR AND BIKE CAN TRAVEL FASTER THAN THE CYCLE BUT THEY ALL WILL HAVE TO TRAVEL WITH THE SPEDD OF THE CYCLE. SIMILAR IS THE CASE WID ELECTRONS. U SAID THT THE RESISTABCE OF R1 IS GREATER .OK THEN IT WILL HINDER THE FLOW OF ELECTRONS . NOW THE AMOUNT OF CHARGE FLOWING THROUGH R1 PER SECOND(THT IS CURRENT) IS THE SAME AMOUNT THT WILL HAVE TO FLOW THROUGH R2 COZ. ONLY THT AMOUNT IS ALLOWED TO PASS THROUGH R1. THTS WHY WE CAN SAY THT THE CURRENT IS SAME IN CASE OF SERIES COMBINATION.
HOPE U GOT UR ANSWER NOW.

RATE............ IF SATISFIED

sure. this systeem has a finite capacitance but its capacitance is less than tht of a single plate.

@x4
SEE, THIS SYSTEM COULD NOT BE CALLED A CAPACITOR. BECAUSE FIRST OF ALL THE CHARGE IS ONLY GIVEN TO ONE PLATE AND AT THE SECOND PLATE THE CHARGE IS INDUCED WHICH CAN NEVER BE LIKE CHARGE. NOW IF U WANT A SYSTM LIKE THIS U HAVE TO CHARGE BOTH THE PLATES SIMULTANEOUSLY THEREFORE THIS IS LIKE CHARGING TWO DIFFERENT CONDDUCTORS SIMULTANEOUSLY.
NOW AS FAR AS UR DERIVATION GOES ON, U MUST FIRST OF ALL KEEP IN CONSIDERATION THE FACT OF USING THE FORMULA
C = Q/V
IN THIS V IS NOT THE POTENTIAL DIFFERENCE BUT THE ABSOLUTE POTENTIAL OF THE CONDUCTOR. NOW KEEPING THIS FACT IN MIND U CAN SEE THT DUE TO THE PRESENCE OF OTHER SIMILARLY CHARGED PLATE THE POTENTIAL AT FIRST PLATE IS ACTUALLY INCREASED SINCE POTENTIAL IS A SCALER QUANTITY AND THEREFORE ALGEBRAIC SUM WILL BE DONE. THEREFORE DUE TO INCREASE IN POTENTIAL THE ACTUAL CAPCITY IS DECREASED NOT INCREASED.

I THINK THIS GOES RIGHT.

PLZ... RATE IF SATISFIED................

SEE, AT LOWER TEMP.WHEN CURRENT STARTS FLOWING THE HEAT IS PRODUCED BUT VERY SOON THE RATE OF PRODUCTION OF HEAT BECOMES EQUAL TO RATE OF DISSIPATION. THERE, FORE VERY SOON THE TEMP. BECOMES CONSTANT AND HENCE IT IS SAID THT
V = IR
BUT AT HIGHER TEMP. THE RISE IN TEMP. IS CONDIDERABLE SINCE HEAT PRODUCED IS VERY LARGE AND HENCE RESISTANCE CHANGES CONTINUOUSLY CHANGING CURRENT AND HENCE RISE IN TEMPERATURE.N SO, IS THE CASE OF NON-OHMIC RESISTORS.
THERE FORE AT LOWER TEMP., V = IR HOLDS GOOD.

RATE IF SATISFIED OR CORRECT ME.

QUESTION 2 - I THINK THE OPTION (B) IS CORRECT. THE ENERGY SUPPLIED SHOULD BE

SOLUTION.- SINCE THE CHARGE Q0 IS REVOLVING AROUND THE CHARGE Q THERE FORE , BY THE CONCEPT OF CENTRIPETAL FORCE

NOW FROM THIS U WILL GET

CALCULATING THE KINETIC ENERGY U WILL GET

(NEGATIVE SIGN IS TAKEN BECOZ. THE FORCE IS ATTRACTIVE)

AND THE TOTAL ENERGY WILL BE

E = KE + PE

THUS,

DOING THE SAME PROCEDURE TAKING RADIUS 2R U WILL GET THE ENERGY AS

NOW CALCULATE ENERGY CHANGE THT IS

E = E2 - E1

AND U WILL GET

THIS IS THE REQUIRED ANSWER AND HENCE OPTON (B) IS CORRECT

HOPE U GOT IT.

PLZ.... RATE IF SATISFIED

@aadit.
SEE , FOR THIS TYPE OF QUESTION WHEN WE HAVE A NEUTRAL CONDUCTING PLATE , THEN TO FIND THE FORCE DUE TO INDUCED CHARGE, WE USE THE IMAGE METHOD. CONSIDER THE MIRROR IMAGE OF THE CHARGE ABT THE PLATE AND FIND THE FORCE DUE TO THT CHARGE ON THE CHARGE CONSIDERED. THIS IS HOW WE CAN FIND THE FORCE ON THE FIRST CHARGE.

SOLVE IT FOR URSELF AND U WILL GET THE EXPRESSION I HAVE USED IN MY SOLUTION.

RATE IF SATISFIED.

I THINK IT SHOULD BE SOFT IRON BECOZ. IT HAS NARROW HYSTERESIS LOOP SO THT ITS AREA AND HENCE THE HEAT PRODUCED DUE TO REPEATED MAGNETISATION AND DEMAGNETISATION IS LESS AND SECOND DUE TO ITS LOWER RETENTIVITY SO THT IT CAN BE DEMAGNETISED EASILY BY APPLYING LOW REVERSE FIELD.

RATE IF SATISFIED .

SEE,

THE FORCE EXPERIENCED BY ANY CHARGE KEPT x DISTANCE AWAY FROM A CONDUCTING PLATE IS GIVEN BY

NOW, SUPPOSE THIS PARTICLE IS MOVED BY A SMALL DISTANCE dx.

THEREFORE, THE WORK DONE WILL BE

NOW INTEGRATE THIS FROM L TO INFINITY THT IS

AND U WILL GET THE ANSWER AS

I THINK THIS IS THE REQUIRED ANSWER AND PROBABLE SOLUTION.

RATE IF SATISFIED OR U CAN ASK FOR FURTHER EXPLANATION.

PLZ.... RATE