Question

A square insulating sheet 70.0 cm on a side is held horizontally. The sheet has 3.50 nC of charge spread uniformly over its area. Part A Calculate the magnitude of the electric field at a point 0.100 mm above the center of the sheet. Express your answer with the appropriate units. E E = nothing nothing Request Answer Part B Estimate the magnitude of the electric field at a point located a distance 200 m above the center of the sheet. Estimate the magnitude of the electric field at a point located a distance 200 above the center of the sheet. E = 2.25×10−5 N/C E = 9.83×10−5 N/C E = 7.87×10−4 N/C E = 1.10×10−2 N/C Request Answer Part C Would the answers to parts A and B be different if the sheet were made of a conducting material? Select the correct answer and explanation. Would the answers to parts and be different if the sheet were made of a conducting material? Select the correct answer and explanation. The charge would automatically spread out evenly over both faces, giving it half the charge density on either face as the insulator and the same electric field only close to the sheet. The answer to part A would not change, but the answer to part B would change. The charge would automatically spread out evenly over both faces, giving it half the charge density on either face as the insulator and changing the sign of the electric field. Both answers would change. The charge would automatically spread out evenly over both faces, giving it half the charge density on either face as the insulator and changing the electric field. Far away, they both look like points with the same charge. The answer to part B would not change, but the answer to part A would change. The charge would automatically spread out evenly over both faces, giving it half the charge density on either face as the insulator but the same electric field. Far away, they both look like points with the same charge. Neither answer would change.

Answer 1

Part A --1) as we know from the given information, the distance of the the point is very small as compared to the sheet, the sheet will act as an infinite sheet off charge.

The electric field produced due to sheet = E = charge density / 2 * ε₀ ( ε₀ = epsilon which has the value as ε₀ =8.85×10⁻¹²)

as charge density = 3.50*10^-9

So, E = 3.50*10^-9 / (2*8.85×10⁻¹² *0.70 *0.70)

E = 4.0355 *10² N/C

Therefore, the electric field at a point 0.100mm above the center of the sheet = 4.0355*10² N/C

Part B :

Now the sheet and the point can be as two point charge where the other point charge is located at a distance of 200 m above the center of the sheet.

so, E = k*q/r²

where, k = coulomb's law

q= charge

r =distance between the two charges

E =9*10⁹*3.50*10^-9/200*200

=7.875*10^-5 N/C

Part C : The Answers to Part A and Part B will not change if the sheets were made of a conducting material.

The charge would automatically spread out evenly over both faces, giving it half the charge density on either face.