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Magnetic Field Due to a Circular Arc at the Centre

Three concetr...

Question

Three concetric spherical conductor A, B and C of radii R, 2R and 4R respectively. A and C is shorted and B is uniformly charged (charge +Q)

Charge on conductor A is

$\frac{Q}{3}$

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$\frac{- Q}{3}$

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$\frac{2 Q}{3}$

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None of these

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Solution

The correct option is B
$\frac{- Q}{3}$

$q_{1} + q_{2} = 0 v_{A} = \frac{k q_{1}}{R} + \frac{k Q}{2 R} + \frac{k q_{2}}{4 R} v_{C} = \frac{k q_{1}}{4 R} + \frac{k Q}{4 R} + \frac{k q_{2}}{4 R} v_{A} = v_{C} \Rightarrow q_{1} = \frac{- Q}{3} and q_{2} = \frac{Q}{3}$

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Three concetric spherical conductor A, B and C of radii R, 2R and 4R respectively. A and C is shorted and B is uniformly charged (charge +Q) Charge on conductor A is

The correct option is B −Q3 q1+q2=0vA=kq1R+kQ2R+kq24RvC=kq14R+kQ4R+kq24RvA=vC⇒ q1=−Q3 and q2=Q3

Three concetric spherical conductor A, B and C of radii R, 2R and 4R respectively. A and C is shorted and B is uniformly charged (charge +Q)

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$\frac{- Q}{3}$

$q_{1} + q_{2} = 0 v_{A} = \frac{k q_{1}}{R} + \frac{k Q}{2 R} + \frac{k q_{2}}{4 R} v_{C} = \frac{k q_{1}}{4 R} + \frac{k Q}{4 R} + \frac{k q_{2}}{4 R} v_{A} = v_{C} \Rightarrow q_{1} = \frac{- Q}{3} and q_{2} = \frac{Q}{3}$