Question

1.

A positive charge of 4.60 C is fixed in place. From a distance of 3.90 cm a particle of mass 6.30 g and charge +3.50 C is fir

2.

The electric potential at a position located a distance of 18.9 mm from a positive point charge of 7.90x10-9C and 14.5 mm fro

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Answer #1

Answer-I Let q, is fixed charge of a & 92 Let E, 2, is fired towards given: 9,= + 4.60 4c = +4.60 x 10°C 2, = + 3.50 4C = + 311 Now we have to find the closest distances that 22 will attain from e, , and Ez starts travelling bock initially the system7 9 -12 8.987X10 X 4.60 X 3.50X1O - we ..9,. . 8.987 X 10 X 4.60x 3. Soxlo X6.30 X10 3.90x10? 3 => 19164.6 x 10 + 37.100 18x1- Ans=2 given v= 1026 KV = 1026 x 16v (di) distance from 2. = 18.9 mm 9,= 7.90 x 10 - 9 distance from 22, do 14.5 mm single121 3 = 3 1.26x10 = 3.7564709x10 + 10.6197931X10 ) 22 3. 3 => 60.6197931x16?) 22 = - 7:4964709x16 = 22= - 204964709 x 10 0.6so answers are- 02= 6.32 5 382 x 10m or Timm = lo ml E = 6.32 5382 mm 08 Leo I cma lo m7 r= 0.6325382 com -37 10 m m 2 = - 4

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