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Q.

In a uniform vertical electric field of intensity E = 20 kV / m, we put a ball of mass m = 10 g and charge with $q = 10 μ C$ (see figure) attached to a light thread of length $l = 40 c m .$ The ball was deflected from the equilibrium position (vertical) to an angle $α = 60 °$ and released from rest. Find the tension (in N) in the thread when the ball passes the equilibrium position. Field lines are vertical. The value of 10T in newton is $(g = 10 m . s^{- 2})$

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Detailed Solution

$(E q - m g) l (1 - c o s 60) = \frac{1}{2} m v^{2}$

$l [2 \times 10^{4} \times 10^{- 5} - 10^{- 2} \times 10] (\frac{1}{2}) = \frac{1}{2} m v^{2}$

$\Rightarrow \frac{m v^{2}}{l} = 0.1$

$T = [E q - m g] + \frac{m v^{2}}{l} = (0.1) + 0.1 = 0.2 N$

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In a uniform vertical electric field of intensity E = 20 kV / m, we put a ball of mass m = 10 g and charge with q=10μC (see figure) attached to a light thread of length l=40 cm. The ball was deflected from the equilibrium position (vertical) to an angle α=60° and released from rest. Find the tension (in N) in the thread when the ball passes the equilibrium position. Field lines are vertical. The value of 10T in newton is (g=10m.s−2)