A charge 17.7 times 10^{-4} C is distributed | vedclass

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Question

(A) The surface charge density $\sigma$ is given by the ratio of charge $q$ to area $A$: $\sigma = \frac{q}{A} = \frac{17.7 	imes 10^{-4} \ C}{200 \

Vedclass · Accepted Answer

$5 	imes 10^5 \ N/C$

Vedclass · Answer

(A) The surface charge density $\sigma$ is given by the ratio of charge $q$ to area $A$: $\sigma = \frac{q}{A} = \frac{17.7 	imes 10^{-4} \ C}{200 \ m^2} = 8.85 	imes 10^{-6} \ C/m^2$. For a large non-conducting sheet,the electric field intensity $E$ at any point near it is independent of the distance and is given by the formula: $E = \frac{\sigma}{2\varepsilon_0}$. Substituting the values: $E = \frac{8.85 	imes 10^{-6}}{2 	imes 8.85 	imes 10^{-12}} = \frac{10^{-6}}{2 	imes 10^{-12}} = 0.5 	imes 10^6 = 5 	imes 10^5 \ N/C$. Thus,the correct option is $A$.

$A$ charge $17.7 \times 10^{-4} \ C$ is distributed uniformly over a large sheet of area $200 \ m^2$. The electric field intensity at a distance $20 \ cm$ from it in air will be $\left[\varepsilon_0=8.85 \times 10^{-12} \ C^2/Nm^2\right]$

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