A charged ball of mass 5.88 × × 10 -4 kg is suspended from two silk strings of equal lengths so that the strings are inclined at 90° with each other. Another ball carrying a charge which is equal in magnitude but opposite in sign of the first one is placed vertically below the first one at a distance of 4.2 × × 10 -2 m. Due to this, the tension in the strings is doubled. Determine the charge on the ball and the tension in the strings after electrostatic interaction. (g = 9.8 m/s 2 )
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(3.36 × 10 -8 C, 8.15 x 10 -3 N)
Sol.
2T 0 cos45 0 = mg
T 0 = 
4T 0 cos45 0 = mg + 
⇒ 2 
T 0 = mg + 
∴ mg = 
or Q = d 
= 4.2 × 10 –2 
= 3.36 × 10 –8 C
Now, T 0 = 
= 
= 4.075 × 10 –3 N
∴ 2T 0 = 8.15 × 10 –3 N
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