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(D) The maximum charge $Q_{	ext{max}}$ that a spherical conductor of radius $R$ can hold in an external electric field $E$ before the air around it b

Vedclass · Accepted Answer

$0.02$

Vedclass · Answer

(D) The maximum charge $Q_{	ext{max}}$ that a spherical conductor of radius $R$ can hold in an external electric field $E$ before the air around it breaks down is given by the condition where the electric field at the surface due to the charge equals the breakdown field strength. However,in this context,the formula for the induced charge capacity is $Q_{	ext{max}} = 4 \pi \varepsilon_0 R^2 E$.Given:Diameter $d = 6 \ mm$,so radius $R = 3 \ mm = 3 	imes 10^{-3} \ m$.Electric field $E = 2 	imes 10^7 \ N/C$.Constant $\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 \ N \cdot m^2/C^2$.Substituting the values:$Q_{	ext{max}} = \frac{1}{9 	imes 10^9} 	imes (3 	imes 10^{-3})^2 	imes (2 	imes 10^7)$$Q_{	ext{max}} = \frac{1}{9 	imes 10^9} 	imes (9 	imes 10^{-6}) 	imes (2 	imes 10^7)$$Q_{	ext{max}} = 10^{-15} 	imes 2 	imes 10^7 = 2 	imes 10^{-8} \ C$$Q_{	ext{max}} = 0.02 	imes 10^{-6} \ C = 0.02 \ \mu C$.

$A$ spherical conductor of diameter $6 \ mm$ is kept in a uniform electric field of intensity $2 \times 10^7 \ N/C$. The maximum charge on the conductor is $\left[\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}\right]$. (in $\mu C$)

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