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(B) The electric field intensity $E$ at a distance $d$ from the centre of a charged sphere is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \cdot \fra

Vedclass · Accepted Answer

$6.67$

Vedclass · Answer

(B) The electric field intensity $E$ at a distance $d$ from the centre of a charged sphere is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{|q|}{d^{2}}$.Since the electric field points radially inward,the charge $q$ must be negative.Given: $E = 1.5 	imes 10^{3} \; N/C$,$d = 20 \; cm = 0.2 \; m$,and $\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \; Nm^{2}/C^{2}$.Rearranging the formula for $q$: $|q| = E \cdot d^{2} \cdot (4 \pi \varepsilon_{0}) = \frac{E \cdot d^{2}}{9 	imes 10^{9}}$.Substituting the values: $|q| = \frac{1.5 	imes 10^{3} 	imes (0.2)^{2}}{9 	imes 10^{9}} = \frac{1.5 	imes 10^{3} 	imes 0.04}{9 	imes 10^{9}} = \frac{0.06 	imes 10^{3}}{9 	imes 10^{9}} = \frac{60}{9 	imes 10^{9}} \approx 6.67 	imes 10^{-9} \; C$.Thus,the magnitude of the net charge is $6.67 \; nC$.

$A$ conducting sphere of radius $10 \; cm$ has an unknown charge. If the electric field $20 \; cm$ from the centre of the sphere is $1.5 \times 10^{3} \; N/C$ and points radially inward,what is the net charge (in $nC$) on the sphere?

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