Question

A uniform electric field of strength ${10}^{6}V/m$ is directed vertically downwards. A particle of mass $0.01kg$ and charge ${10}^{-6}$ coulomb is suspended by an inextensible thread of length $1m$. The particle is displayed slightly from its mean position and released.

(a) Calculate the time period of its oscillation.

(b) What minimum velocity should be given to the particles at rest so that it completes a full circle in a vertical plane without the thread getting slack?

(c) Calculate the maximum and minimum tensions in the thread in this situation.

• A. $0.6$sec (b) $23.42m/s$ (c) $9.588N$, Zero
• B. $2.6$sec (b) $20.42m/s$ (c) $3.588N$, Zero
• C. $1.6$sec (b) $21.42m/s$ (c) $5.588N$, Zero
• D. $0.6$sec (b) $23.42m/s$ (c) $6.588N$, Zero

Answer 1

Answer: (D)

$0.6$sec (b) $23.42m/s$ (c) $6.588N$, Zero

A uniform electric field at strength $={10}^{6}V/m$

$m=0.1kg$

$Q={10}^{-6}C$

$L=1m$

(a) $T=2\pi \sqrt{\frac{m}{g}}$ $=2\pi \sqrt{\frac{0.01}{10}}=0.6$ sec

(b) minimum velocity $V=\sqrt{2gh}=\sqrt{2gl/q}=\sqrt{2\times 10\times 1/{10}^{-6}}=23.42m$

(c) Tension of maximum $T={10}^1\times 0.01\times 0.6=6.588N$

minimum tension $=0$