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CBSE Class 12-science Answered

A pendulam bob of mass 80 x 10 -6 kg carrying a charge of 2 x10 -8 C is at rest in a horizontal uniform electric field of 2 x 10 4 NC -1 . Find the tension in the thread of the pendulam and the angle it makes with the verticle .
Asked by rutujasarangmehta | 04 Mar, 2016, 10:21: AM
answered-by-expert Expert Answer
 
 
begin mathsize 12px style Let space the space tension space in space thread space be space straight T space and space the space thread space make space an space angle space of space straight theta space with space the space vertical. There space will space be space 3 space forces space acting space on space the space bob colon 1 right parenthesis space Wieght space mg space of space the space bob space acting space vertically space downwards. 2 right parenthesis space Tension space straight T space along space AO 3 right parenthesis space Electric space feild space QE space which space is space acing space in space the space horizontal space direction The space bob space is space at space equilibrium. space From space the space diagram comma space we space get straight T space sinθ space equals space QE space and space straight T space cosθ space equals space mg therefore space tan straight theta space equals space QE over mg space equals space fraction numerator open parentheses 2 cross times 10 to the power of negative 8 end exponent close parentheses space cross times space open parentheses 2 cross times 10 to the power of 4 close parentheses over denominator open parentheses 80 space cross times space 10 to the power of negative 6 end exponent close parentheses space cross times space 9.8 end fraction therefore space tanθ space equals 0.51 space or space straight theta space equals space tan to the power of negative 1 end exponent left parenthesis 0.51 right parenthesis space equals space 27 degree Also comma space straight T space equals QE over sinθ space equals space fraction numerator open parentheses 2 cross times 10 to the power of negative 8 end exponent close parentheses space cross times space open parentheses 2 cross times 10 to the power of 4 close parentheses over denominator sin 27 degree end fraction therefore space straight T space equals 8.81 space cross times space 10 to the power of negative 4 end exponent space straight N end style
Answered by Yashvanti Jain | 04 Mar, 2016, 11:07: AM

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