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A charged ball B hangs from a silk thread S which makes an angle \[\theta \] with a large charged conducting sheet P as shown in the given figure. The surface charge density \[\sigma \] of the sheet is proportional to:
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A. \[\cos \theta \]
B. \[\cot \theta \]
C. \[\sin \theta \]
D. \[\tan \theta \]

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Last updated date: 14th Jun 2024
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Answer
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Hint: Draw a free body diagram indicating the forces acting on the charged ball. Equate the forces and use the relation between electric field and permittivity of the medium.

Formula used:
\[{F_e} = qE\]
Here, q is the charge of on the ball and E is the electric field.

Complete step by step answer:
The charged ball is placed in a uniform electric field experiences an electric force of magnitude,
\[{F_e} = qE\]
Here, q is the charge of on the ball and E is the electric field.Draw a free body diagram of the forces acting on the charged ball as shown in the figure below,

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Balance the forces acting on the charged ball in the vertical direction as follows,
\[T\cos \theta = mg\] ……. (1)
Balance the forces acting on the charged ball in the horizontal direction as follows,
\[T\sin \theta = {F_e}\]
\[ \Rightarrow T\sin \theta = qE\] ……. (2)
Divide equation (2) by equation (1). \[\dfrac{{T\sin \theta }}{{T\cos \theta }} = \dfrac{{qE}}{{mg}}\]
\[ \Rightarrow \tan \theta = \left( {\dfrac{q}{{mg}}} \right)E\] ……. (3)
We know that the electric field is the ratio of surface charge density \[\sigma \]and permittivity of the medium \[\varepsilon \].
\[E = \dfrac{\sigma }{\varepsilon }\]
Substitute \[E = \dfrac{\sigma }{\varepsilon }\] in equation (3).
\[\left( {\dfrac{q}{{mg}}} \right)\dfrac{\sigma }{\varepsilon } = \tan \theta \]
\[ \Rightarrow \sigma = \left( {\dfrac{{\varepsilon mg}}{q}} \right)\tan \theta \]
\[\therefore \sigma \propto \tan \theta \]

So, the correct answer is “Option D”.

Note:
Specify the correct directions of forces acting on the charged ball whether they are along the positive direction of the y-axis or along the negative direction of the y-axis and the same for the x-axis. Here, the electric force should be along the positive direction of the x-axis.