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Order of Magnitude

A spherical c...

Question

A spherical conductor A contains two spherical cavities. The total charge on the conductor itself is zero. However, there is a point charge $q_{b}$ at the centre of one cavity and $q_{c}$ at the centre of the other. A considerable distance r away from the centre of the spherical conductor, there is another charge $q_{b}$ . Force acting on $q_{b}, q_{c}$ and $q_{d}$ are $F_{1}, F_{2}$ and $F_{3}$ respectively. [Assume all charges are positive]

F_{1} < F_{2} < F_{3}

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F_{1} = F_{2} < F_{3}

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F_{1} = F_{2} > F_{3}

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F_{1} > F_{2} > F_{3}

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Solution

The correct option is B $F_{1} = F_{2} < F_{3}$
Given, spherical conductor A. It contains two spherical cavities. There is a point charge $q_{b}$ at the centre of one cavity and $q_{c}$ at the centre of the other.
An another charge $q_{d}$ is placed at distance $r$ from the centre of the spherical conductor. So, according to problem the figure can be considered as
$∵ E$ inside the cavity is zero at steady state.
$∵ F = \frac{E}{q} \Rightarrow F = 0$
$∴ F_{1} = F_{2} = 0$
Now, from Coulomb's law
$F = \frac{k q_{1} q_{2}}{r^{2}}$
Here $q_{1} = q_{b} + q_{c}, q_{2} = q_{d}$
$F_{3} = \frac{k (q_{b} + q_{c}) q_{d}}{r^{2}}$
$F_{1} = F_{2} < F_{3}$

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