Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

??

qsnImg
Asked by SanskarAgarwal86 2nd June 2019, 12:44 AM
Answered by Expert
Answer:
The direction of the field inside the uniformly charged sphere is radially outwards. Therefore if bullet managed to reach just little bit beyond the center, the electric field will push the bullet outwards. So we will calculate the velocity required to reach the centre.
So by energy conservation
 
K E subscript i n i t i a l end subscript plus P E subscript s u r f a c e end subscript equals K E subscript c e n t e r end subscript plus P E subscript c e n t r e end subscript
1 half m u squared plus fraction numerator K q squared over denominator R end fraction equals 0 plus fraction numerator begin display style 3 K q squared end style over denominator begin display style 2 R end style end fraction
fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction m u squared equals fraction numerator begin display style K q squared end style over denominator begin display style 2 R end style end fraction
u equals square root of fraction numerator q squared over denominator 4 pi epsilon subscript 0 m R end fraction end root
u equals fraction numerator begin display style q end style over denominator begin display style square root of 4 pi epsilon subscript 0 m R end root end style end fraction
Answered by Expert 2nd June 2019, 12:40 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer 4/10

Your answer has been posted successfully!

Chat with us on WhatsApp