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two identical balls each having a charge of 2×10-7 and a mass of 100g are suspended from a common point by two insulating strings each 50 cms long. The balls are held at a separation 5 cm apart and then released . find the 1]the electric force on one of the charged balls.2]the components of the resultant force on it along and perpendicular to the string.3]the tension in the string4]the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release
Hi Find the angle formed by the strings by using sine law (trigonometry) : a /sin a = b/sinb = c/sinc . a=5cm , b= c=50 cm , angle b = angle c 1. u can find the force on the ball by simply using the coulombs law f =kq1q2/r2 where r = 5cm . 2. now since u know the angles, find the resultants in the respective directions by taking components of the force in that direction 3. Draw force body diagram of the ball and balance the forces along the string because it is unstretchable, hence u can find the tension in the string at that instant 4. after that , calaculate acceleration perpendicular to the string by taking into account the components of tension , wt and eletrostatic force in that direction,.
Hi
Find the angle formed by the strings by using sine law (trigonometry) : a /sin a = b/sinb = c/sinc . a=5cm , b= c=50 cm , angle b = angle c
1. u can find the force on the ball by simply using the coulombs law f =kq1q2/r2 where r = 5cm .
2. now since u know the angles, find the resultants in the respective directions by taking components of the force in that direction
3. Draw force body diagram of the ball and balance the forces along the string because it is unstretchable, hence u can find the tension in the string at that instant
4. after that , calaculate acceleration perpendicular to the string by taking into account the components of tension , wt and eletrostatic force in that direction,.
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