Question

A square insulating sheet 90.0 cm on a side is held horizontally. The sheet has 8.50 nC of charge spread uniformly over its area.

1.Calculate the magnitude of the electric field at a point 0.100 mm above the center of the sheet.

2.Estimate the magnitude of the electric field at a point located a distance 200 m above the center of the sheet.

E = 5.46×10−5 N/C
E = 2.39×10−4 N/C
E = 1.91×10−3 N/C
E = 2.68×10−2 N/

3. 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.

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.

At distance of 0.100 mm, sheet will be considered infinite since length of side >>> 0.100 mm, So

Electric field due to a infinite charged sheet is given by:

E1 =

charge density per unit area = Q/A

E1 = Q/(2*e0*A)

Q = charge = 8.50 nC = 8.50*10^-9 C

e0 = 8.85*10^-12

A = Area of square sheet = side^2 = 90 cm*90 cm = 0.81 m^2

So,

E1 = 8.50*10^-9/(2*8.85*10^-12*0.81)

E1 = 592.87 N/C

Part B.

at this point, charged sheet will behave like point charge, So Electric field due to a point charge is given by:

E2 = k*Q/R^2

Q = 8.50 nC

R = 200 m

So,

E2 = 9*10^9*8.50*10^-9/200^2

E2 = 1.91*10^-3 N/C

Part C.

Correct option is D. there would be no difference in both scenario since charge will spread out evenly over both faces, and still be infinite for very near point and point charge for very far point

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