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CBSE Class 12-science Answered

1. The focal length of an equiconvex lens in air is equal to either of its radii of curvature. the refractive index of the material of lens is __________
2. A charge of 4x10^-8C is uniformly distributed over the surface of sphere radius 1cm. another hollow sphere of radius 5cm is concentric with the small sphere. find intensity of electric field at a distance 2cm from centre. [K=9x10^9 MKS] [ans in NC^-1]
3. A light of 4560angstrom and 1mW is made incident on photosensitive surface of Cs. If efficiency of surface is 0.1%, then photoelectric current produces is___________
4.A spherical drop of water has 3x10^-10C charge on its surface. 300 volt electric potential exists on its surface. if 27 such identical drops are combined to form a single drop, calculate electric potential on the surface of new drop._____________
Asked by Dhriti | 24 Oct, 2014, 12:10: PM
answered-by-expert Expert Answer
2. 
 
Given charge Q = 4 × 10-8 C. 
begin mathsize 14px style Let space straight P space be space straight a space point space at space straight a space distance space of space straight r space left parenthesis 2 space cm right parenthesis space away space from space the space centre. To space find space the space field space at space straight P space let space us space consider space straight a space gaussian space surface space through space the space point space straight P space as space shown space in space the space figure. We space can space write space that space the space flux space through space this space surface space equals space contour integral space straight E with rightwards arrow on top. dS with rightwards arrow on top space space space space space space equals space contour integral straight E. space dS space space space space space space equals straight E contour integral dS space space space space space space equals space 4 cross times straight pi cross times straight r squared cross times straight E Here space straight r space equals space 2 space cm space equals space 2 space cross times 10 to the power of minus 2 space end exponent straight m According space t o space Gauss space Law comma space this space flux space is space equal space to space the space charge space straight Q space enclosed space by space the space surface space divided space by space straight epsilon subscript 0. straight i. straight e. space space contour integral space straight E with rightwards arrow on top. dS with rightwards arrow on top space equals space straight Q over straight epsilon subscript 0 4 cross times straight pi cross times straight r squared cross times straight E space equals straight Q over straight epsilon subscript 0 4 cross times straight pi cross times left parenthesis 2 cross times 10 to the power of minus 2 end exponent right parenthesis space squared cross times straight E space equals straight Q over straight epsilon subscript 0 On space rearranging space we space get space : straight E space equals space fraction numerator straight Q over denominator straight epsilon subscript 0 cross times 4 cross times straight pi cross times left parenthesis 2 cross times 10 to the power of minus 2 end exponent right parenthesis squared space end fraction open parentheses we space know space that space fraction numerator 1 over denominator 4 πε subscript 0 end fraction equals k equals 9 cross times 10 to the power of 9 space Nm squared straight C to the power of minus 2 end exponent space space close parentheses space space space space equals 9 cross times 10 to the power of 9 space Nm squared straight C to the power of minus 2 end exponent cross times fraction numerator 4 cross times 10 to the power of minus 8 end exponent space straight C over denominator 4 cross times 10 to the power of minus 4 end exponent space straight m squared end fraction straight E space equals 9 space cross times space 10 to the power of 5 space straight N space straight C to the power of minus 1 end exponent end style
The Electric field at a distance 2cm from centre = 9 × 105 N C-1
 
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