H C VERMA ch-29 Q-22
Asked by Ashish Kumar | 27th Apr, 2013, 08:28: AM
a) The charge on each ball = Q= 2x10-7
Distance between the balls = 5cm
Electrical Force = Q2/4??0 = 4x10-14 / 4x3.1452x8.854x10-12 = 3.6x10-4 N
b) Lets suppose that the angle made by the string and the vertical line is x.
Then sin x = 5/2x50 = 1/20
so cos x = 0.9987
Since the tension in the string will balance the force along the string, The net force along the string will be zero.
The force perpendicular to the string will be
F= mgsinx - Fecosx = 0.1x9.8x0.05 - 0.9987x3.6x10-4 = 0.0486 N
c) The tension in the string will be
T = Fesinx + mgcosx = 3.6x10-4x0.05 + 0.1x9.8x0.9987 = 0.9787 N
d) acceleration of the ball = Net force / mass = 0.0486/0.1 = 0.486 m/s/s
Since the net force is along perpendicular to the string.
Answered by | 8th May, 2013, 10:37: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number