189k views
0 votes
Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charge of uniform density: −4 · 10−5 C/m2 on A, 7 · 10−5 C/m2 on B, and −3 · 10−5 C/m2 on C. (The density given includes charge on both sides of the sheet.) What is the magnitude of the electrical force per unit area on each sheet? Check to see that the total force per unit area on the three sheets is zero.

1 Answer

3 votes

Answer:

0.

Step-by-step explanation:

To find the electrical force per unit area on each sheet we start defining our variables,


\sigma_A = -4.10^(-5)C/m^2


\sigma_B= -7.10^(-5)C/m^2


\sigma_C = -3.1^(-5)C/m^2

We find the electric field for each one, this formula is given by,


E= (\sigma_i)/(2\epsilon_0)

Substituting each value from the three charged sheets, we have


E_A= (-4.10^(-5)C/m^2)/(2\epsilon_0)(\hat{j})


E_B= (-7.10^(-5)C/m^2)/(2\epsilon_0)(-\hat{j})


E_C= (-3.1^(-5)C/m^2)/(2\epsilon_0)(\hat{j})

The electric field is


E_(NET)= E_A+E_B+E_C


E_(NET)=(-4.10^(-5)C/m^2)/(2\epsilon_0)(\hat{j})+(-7.10^(-5)C/m^2)/(2\epsilon_0)(-\hat{j})+ (-3.1^(-5)C/m^2)/(2\epsilon_0)(\hat{j})


E_(NET) = 0

Force on each sheet is,


F=E_(NET)\sigma ds


F=0

The total force is 0

User Cetcet
by
8.3k points