Two infinite sheets of charge are separated by 10 cm as shown in the gure. Sheet

Question

Two infinite sheets of charge are separated by 10 cm as shown in the gure. Sheet 1 has a surface charge distribution of 1-2.45 pC/m2 and sheet 2 has a surface charge distribution of ,--4.01 C/m 2, Find the x-component of the electric field at each of the following locations: Sheet 1 10 cm a) at point P, 6 cm to the left of sheet 1 Submit Answer Incorrect. Tries 5/10 Previous Tries 6 cm-6 cm- b) at point P, 6 cm to the right of sheet 1 Submit Answer Incorrect. Tries 2/10 Previous Tries This discussion is closed

Explanation / Answer

The most general expression for electric field intensity due to charged sheet is
/2 0.

0 permittivity of free space or air.
One more important point: Field intensity is independent of distance of point.
As electric field intensity is a vector quantity, the direction will be away in case of positive charged sheet and towards if sheet is negatively charged.

(b)
So in our case if P' is in between two sheets, E1 = +2.45*10-6 / 2 0
Due to other sheet, E2 = +4.01*10-6 / 2 0
Here because of negative charge it will also be in the same direction as that of E1.
So we have to sum up. Hence field at P' ie in between two sheets, 6.46*10-6 / 2 0

Field at P' = 6.46*10-6 /(2*8.852*10-12) = 0.364*106

(a)
Now if P is outside of the two sheets, then E1 and E2 will be opposite to each other. So any way difference is needed. So E at a point outside, 1.56*10-6 / 2 0
0 = 8.852*10-12

Field at P = 1.56*10-6 /(2*8.852*10-12) = 0.088*106

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