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Question: A pendulum bob of mass 80 mg and carrying a charge of 2 × 10⁻⁸ \[coul\]is at rest in a horizontal un...

A pendulum bob of mass 80 mg and carrying a charge of 2 × 10⁻⁸ coulcoulis at rest in a horizontal uniform electric field of 20,000 V m⁻1. Find the tension in the thread of the pendulum.
A.8.8 × 10⁻2 N
B.8.8 × 10⁻3 N
C.8.8 × 10⁻4 N
D.8.8 ×10⁻ 5 N

Explanation

Solution

The pendulum is in a rest position. All the forces i.e. components of force and the other force will be in balanced state. When we equate the horizontal and vertical forces, we can find the tension in the thread of a string.

Step by step answer: A pendulum bob is at a rest position in a uniform horizontal electric field E = 20000 Vm-1.. The mass m of bob is 80 mg i.e., 80 × 10-3 kg (1g =10-3kg) is carrying a charge q = 2 × 10⁻⁸ C and let the tension in the thread of the string be T and the angle made by the string with vertical be Ɵ. As the pendulum bob is in a rest position i.e. in equilibrium. So, all the forces acting on the bob will be in a balanced state.
Tsin⁡θ=qET\sin \theta = qE [T sin Ɵ is the component of T acting horizontally which balances the force acted due to electric field]
Tcos⁡θ=mgT\cos \theta = mg [T cos Ɵ is the component of T acting vertically upwards which is balanced by the weight of the body mg]
When we divide the above two equations,
Tsin⁡θTcos⁡θ=qEmg\dfrac{{T\sin \theta }}{{T\cos \theta }} = \dfrac{{qE}}{{mg}}
⇒\Rightarrow tan⁡θ=qEmg[tan⁡θ=sin⁡θcos⁡θ]\tan \theta = \dfrac{{qE}}{{mg}}\left[ {\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}} \right]
⇒\Rightarrow tan⁡θ=2×10−5×2000080×10−3×10\tan \theta = \dfrac{{2 \times {{10}^{ - 5}} \times 20000}}{{80 \times {{10}^{ - 3}} \times 10}}
⇒\Rightarrow tan⁡θ=0.5\tan \theta = 0.5
⇒\Rightarrow tan⁡−1(0.5){\tan ^{ - 1}}\left( {0.5} \right)
⇒\Rightarrow θ=27[tan⁡−1(0.5)=26.5]\theta = 27\left[ {{{\tan }^{ - 1}}\left( {0.5} \right) = 26.5} \right]
Now, from the first equation ,
Tsin⁡θ=qE ⇒T=qEsin⁡θ ⇒T=2×10−5×200000.453 ⇒T=88300.22×10−8  T\sin \theta = qE \\\ \Rightarrow T = \dfrac{{qE}}{{\sin \theta }} \\\ \Rightarrow T = \dfrac{{2 \times {{10}^{ - 5}} \times 20000}}{{0.453}} \\\ \Rightarrow T = 88300.22 \times {10^{ - 8}} \\\ \thereforeT=8.8×10−4 T = 8.8 \times {10^{ - 4}} \\\
Therefore, option C is correct.

Note: The electric field acts in a horizontal direction which balances the horizontal component of force and the weight of the bob balances the vertical component of force. In equilibrium, all the forces are in a balanced state.