A spherical conductor of radius 4 cm carries a charge of 1-11- 10-10 C- At distance 4D in cm from centre of sphere a point charge of 4-44- 10-10 C is placed such that the potential of the conductor becomes 50 V- Assume V-0 at - Find D

1/4 πϵQq/4D = 50 ⇒ D=1/4 πϵQq/4× 50=2.21576 × 10^ 12 m = nbsp;=2.21576 × 10^ 10 cm nbsp; nbsp; nbsp; ...

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Standard XII

Physics

Charges on Conductors

A spherical c...

Question

A spherical conductor of radius 4 cm carries a charge of $1.11 \times 10^{- 10}$ C. At distance 4D (in cm) from centre of sphere a point charge of $4.44 \times 10^{- 10}$ C is placed such that the potential of the conductor becomes 50 V? Assume $V = 0$ at $\infty$ . Find D

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Solution

$\frac{1}{4 π ϵ} \frac{Q q}{4 D} = 50 \Rightarrow D = \frac{1}{4 π ϵ} \frac{Q q}{4 \times 50} = 2.21576 \times 10^{- 12} m == 2.21576 \times 10^{- 10} c m$

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A spherical conductor of radius 4 cm carries a charge of 1.11×10−10 C. At distance 4D (in cm) from centre of sphere a point charge of 4.44×10−10 C is placed such that the potential of the conductor becomes 50 V? Assume V=0 at ∞. Find D

14πϵQq4D=50⇒D=14πϵQq4×50=2.21576×10−12m==2.21576×10−10cm

A spherical conductor of radius 4 cm carries a charge of $1.11 \times 10^{- 10}$ C. At distance 4D (in cm) from centre of sphere a point charge of $4.44 \times 10^{- 10}$ C is placed such that the potential of the conductor becomes 50 V? Assume $V = 0$ at $\infty$ . Find D

$\frac{1}{4 π ϵ} \frac{Q q}{4 D} = 50 \Rightarrow D = \frac{1}{4 π ϵ} \frac{Q q}{4 \times 50} = 2.21576 \times 10^{- 12} m == 2.21576 \times 10^{- 10} c m$