A circular plate sheet of radius 10 ,cm is pl | vedclass

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(B) Given: Radius of the plate, $R = 10 \,cm = 0.1 \,m$.Uniform electric field, $E = 2 \sqrt{3} 	imes 10^5 \,NC^{-1}$.The angle between the plate and

Vedclass · Accepted Answer

$9.42 	imes 10^3 \,Nm^2 C^{-1}$

Vedclass · Answer

(B) Given: Radius of the plate, $R = 10 \,cm = 0.1 \,m$.Uniform electric field, $E = 2 \sqrt{3} 	imes 10^5 \,NC^{-1}$.The angle between the plate and the electric field is $60^{\circ}$.The angle $	heta$ between the normal to the plate and the electric field is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.The area of the circular plate is $A = \pi R^2 = \pi (0.1)^2 = 0.01 \pi \,m^2$.The electric flux $\phi$ is given by the formula $\phi = EA \cos 	heta$.Substituting the values:$\phi = (2 \sqrt{3} 	imes 10^5) 	imes (0.01 \pi) 	imes \cos 30^{\circ}$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, we have:$\phi = 2 \sqrt{3} 	imes 10^5 	imes 0.01 \pi 	imes \frac{\sqrt{3}}{2}$.$\phi = 3 	imes 10^3 	imes \pi = 3 	imes 3.14159 	imes 10^3 = 9.42477 	imes 10^3 \,Nm^2 C^{-1}$.Rounding to two decimal places, $\phi \approx 9.42 	imes 10^3 \,Nm^2 C^{-1}$.

$A$ circular plate sheet of radius $10 \,cm$ is placed in a uniform electric field of $2 \sqrt{3} \times 10^5 \,NC^{-1}$, making an angle of $60^{\circ}$ with the field. Find the electric flux through the sheet.

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