question_answer

A pendulum bob of mass m carrying a charge q is at rest with its string making an angle \[\theta \] with the vertical in a uniform horizontal electric field E. The tension in the siring is

A) \[\frac{\text{mg}}{\text{sin}}\text{and}\frac{\text{qE}}{\text{cos}\,}\]                

B) \[\frac{\text{mg}}{\text{cos}}\text{and}\frac{\text{qE}}{\sin \,}\]           

C) \[\frac{\text{qE}}{mg}\]                              

D)        \[\frac{mg}{\text{qE}}\]

Correct Answer: B

Solution :

A pendulum bob of weight mg and carrying a charge q is suspended in the electric field E. Suppose that it comes to rest at point A, so that it makes an angle \[\theta \] with the vertical as shown. At point A, the bob is acted upon by the following three forces. (i) Weight mg acting vertically downward. (ii) Tension T in the string along AS (iii)Electrostatic force qE on the bob along horizontal. Since, the bob is in equilibrium under the action of the forces mg, T and qE, these forces can be represented by the sides SN, AS and NA of triangle ANS. Therefor,             \[\frac{mg}{SN}=\frac{qE}{NA}=\frac{T}{AS}\] \[\Rightarrow \]               \[qE=T\sin \theta \] and        \[mg=T=\cos \theta \]