Solved: Density of Induced Charge is 7.083*10-7 C/m2

[Dell]

2 parrallel boards are charged with an equal and oposite charge, when the area between them is emptied, the electric field reaches 2*10⁵ V/m. when the area is filled with a dialectric substance, the field is reduced to 1.2*10⁵V/m. what is the density of the induced charge??

this is what i have done so far, could someone tell me if it is correct, and if not where i am going wrong.

E=1.2*10⁵
E₀=2*10⁵

K=E₀/E=5/3

∫Kε₀EdA=Q_free
==>Q_free=ε₀KEA

∫ε₀EdA=Q_free+Q_induced
==>Q_induced=ε₀KEA-ε₀EA

Q_ind/A=ε₀E(1-K)

=ε₀(1.2-2)*10⁵

=7.083*10^-7C/m²

 

[nickjer]

∫ε₀EdA=Q_free+Q_induced
==>Q_induced=ε₀KEA-ε₀EA

Seems you messed up the negative sign here, but you corrected it in the following line. The rest looks good, I believe though that the final charge density should be written as negative since it makes an electric field in the opposite direction to reduce the total field. But that depends on how picky the grader is.

 

[nlightner]

Your approach is correct. However, there are a few minor errors in your calculations. Here is a corrected version:

First, we need to determine the charge density on the parallel boards when they are empty. We can use the given electric field and the formula for electric field, E = σ/ε0, to find the surface charge density, σ, on the boards:

E = 2*105 V/m
σ = E * ε0 = 2*105 * 8.85*10-12 = 1.77*10-6 C/m2

Now, when the area between the boards is filled with a dielectric substance, the electric field is reduced to 1.2*105 V/m. Using the same formula, we can find the new surface charge density:

E = 1.2*105 V/m
σ = E * ε0 = 1.2*105 * 8.85*10-12 = 1.06*10-6 C/m2

The induced charge density is the difference between these two values:

σind = σempty - σfilled
= (1.77*10-6 - 1.06*10-6) C/m2
= 7.083*10-7 C/m2

This matches the given value for the induced charge density, so your approach and calculations are correct.