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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: charge of ball
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sneha_91 (27)

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a system consist a ball of radius "r" carrying spherically symmentric  charge and the surrounding space filled with volume density=/r .alpha=constsnt r is the dis. from the centre of the ball  .


a)find the ball's charge at which the mag of elctricfield vector is independant of r outside the ball


b)how high is the field strength? permitivity of ball and surrounding is 1

    
risin (179)

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Charge=


calculate E with gauss law.

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animal (610)

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hey i m not getting ur question


is the surrounding space filled with charge?

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indrajeet_bariar (245)

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YES SNEHA
IS IT THE VOLUME CHARGE DENSITY OF THE SURROUNDING MEDIUM???
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indrajeet_bariar (245)

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I THINK THE ANSWER THT IS THE CHARGE ON THE SPHERE SHOULD BE

 


                                                        


plz... rate if answer is correct. u can ask further if u require the solution.

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animal (610)

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indrajeet sir plz  tell the soln.

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sneha_91 (27)

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spere has a radius "R"..r is the distance from the center.....this question is in iridov

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sneha_91 (27)

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idrajeet can u explain your solution


 

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indrajeet_bariar (245)

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@sneha

ok now i got the question right.

I THINK NOW I HAVE TO CHANGE THE ANSWER AND PUT A SOLUTION.

TO CALCULATE THE ELECTRIC FIELD AT A POINT OUTSIDE THE SPHERE, DRAW A GAUSSIAN SURFACE THT IS A SPHERE OF RADIUS r THROUGH THE POINT.NOW INSIDE THIS GAUSSIAN SURFACE, THE TOTAL CHARGE WILL BE DUE TO CHARGE ON THE SPHERE WHICH IS LET US ASSUME Q AND OTHER DUE TO THE CHARGED MEDIUM. THE CHARGE DUE TO THE MEDIUM CAN BE CALCULATED AS BELOW.


let us assume a point at a distance r through which there is a sphere of width dr.


now charge dq will be


            


                   


                  


 now,            


 


                    


 


solving this for E we will get


                   


now for electric field to be free from r the second expression inside the brackets must be equal to zero.


                             


 or,                     


i think this is the required charge on the sphere.


hope u understood the solution.


PLZ... VOTE IF SATISFIED WITH THE SOLUTION.


 

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mukundmadhav (460)

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risin is right..
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