Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform

Question

Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform charge densities 1, 2, 3, and 4 on their surfaces, as shown in the figure (Figure 1) . These surface charge densities have the values 1= -5.65 C/m2 , 2=+ 5.45 C/m2 , 3=+ 2.25 C/m2 , and 4=+ 3.85 C/m2 .

A. Find the magnitude of the electric field at the point A, 5.00 cm from the left face of the left-hand sheet.

B. Find the magnitude of the electric field at the point B, 1.25cm from the inner surface of the right-hand sheet.

C. Find the magnitude of the electric field at the point C, in the middle of the right-hand sheet.

Explanation / Answer

Here ,

A)

electric field due to charge sheet , E = sigma/(2*epsilon)

at the point A ,

E = (sigma1 + sigma2 + sigma3 +sigma4)/(2 * epsilon)

E = (-5.65 + 5.45 + 2.25 + 3.85) *10^-6/(2 * 8.854 *10^-12)

E = 3.33 *10^11 N/C

the electric field is 3.33 *10^11 N/C

B)

at the point B ,

E = (sigma1 + sigma2 - sigma3 - sigma4)/(2 * epsilon)

E = (-5.65 + 5.45 - 2.25 - 3.85) *10^-6/(2 * 8.854 *10^-12)

E = -3.557 *10^11 N/C

the electic field at B is -3.557 *10^11 N/C

C)

Now, at point C ,

E = (sigma1 + sigma2 + sigma3 - sigma4)/(2 * epsilon)

E = (-5.65 + 5.45 + 2.25 - 3.85) *10^-6/(2 * 8.854 *10^-12)

E = -1.02 *10^11 N/C

the electric field at C is -1.02 *10^11 N/C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.