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A nonconducting thin layer carries charge with a uniform density of 8.95 µC/cm2.

(a) What is the electric field 7.55 cm in front of the wall if 7.55 cm is small compared with the dimensions of the wall?

magnitude

___ N/C

direction (away from or towards the wall)

(b) Does your result change as the distance from the wall varies? (Assume that the distance from the wall is small compared to the width and height of the wall.)

yes or no

(c*) The nonconducting wall is replaced with a thick conducting wall with the same surface charge density on the right side of the conducting wall as was on the thin insulating layer. What is the electric field 7.55 cm in front of (just outside) the conducting wall if 7.55 cm is small compared with the dimensions of the wall?

magnitude ________N/C

direction? (away from or towards the wall)

Respuesta :

Answer:

5.06*10^9

Explanation:

EA = σ(πr²)/εo, where area of the wall A = 2πr², on substituting we have

E(2πr²) = σ(πr^2)/εo

E = σ/(2εo)

What we then do is substitute the values from the question into this formula

E = (8.95*10^-6 C/cm^2)(100^2 cm^2 / m^2) / (2*8.85e-12 C^2/N-m^2)

E = 5.06*10^9 N/C

Therefore, the electric field is calculated to be 5.06*10^9 N/C

batolisis

A) The electric field in front of the wall is  5.05 * 10⁹ N/C,   Away from the wall

B) No the result does not change as distance from wall varies

C) The electric field in front of the conducting wall is :  10.11 * 10⁹ N/C ,   Away from the wall

Given data :

uniform density of nonconducting thin layer = 8.95 µC/cm² = 8.95 * 10⁻² C/m²

A) Determine the electric field 7.55 cm in front of the wall

E = [tex]\frac{\alpha }{2*\beta }[/tex]  --- ( 1 )

where :  [tex]\alpha[/tex] =  8.95 * 10⁻² C/m²,   [tex]\beta[/tex]= 8.85 * 10⁻¹² N-m²/C²

back to equation ( 1 )

E = ( 8.95 * 10⁻² ) / (  8.85 * 10⁻¹² ) * [tex]\frac{1}{2}[/tex]

  =  5.05 * 10⁹ N/C,

B) The result from part A does not change as the distance from the wall varies. ( No )

C) When the nonconducting wall is replaced by thick conducting wall

Determining the electric field strength for the conducting wall

E = [tex]\frac{\alpha }{\beta }[/tex]  ----- ( 2 )

where : [tex]\alpha[/tex] =  8.95 * 10⁻² C/m²,  [tex]\beta[/tex] = 8.85 * 10⁻¹² N-m²/C²

E = ( 8.95 * 10⁻² ) / ( 8.85 * 10⁻¹² )

  = 10.11 * 10⁹ N/C

Hence we can conclude that;  The electric field in front of the wall is  5.05 * 10⁹ N/C,   Away from the wall.  No the result does not change as distance from wall varies, and The electric field in front of the conducting wall is :  10.11 * 10⁹ N/C ,   Away from the wall.

Learn more about electric field calculations : https://brainly.com/question/14372859

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