How Do You Calculate E in Mica for a Capacitor?

[mr_coffee]

A parallel-plate capacitor has a capacitance of 60 pF, a plate area of 130 cm2, and a mica dielectric ( = 5.4). At 60 V potential difference, calculate the following values.

(a) E in the mica
wrong check mark V/m
(b) the magnitude of the free charge on the plates
C
(c) the magnitude of the induced surface charge on the mica
C

I tried the following:
E = [Q/A]/(k*Eo);
E = (1e-12/.013)/(5.4)(8.85e-12);
E = 1.609 which is wrong.

I got Q by the following: C = QV; Q = C/V; 60e-12/60 = 1e-12

 

[quantumdude]

Sorry for the late reply. If you're still working on this then check out the websites I gave you in this thread and see if you don't have better luck.

 

[quicknote]

;

I would approach this problem by first reviewing the given information and equations related to capacitors and dielectrics. I would also check the units used in the calculations to ensure consistency. Here is my response to the given content:

Firstly, it is unclear what the wrong check mark symbol represents in the given content. Assuming it was intended to represent the units of E, it should be V/m instead of C. Also, the correct units for E are V/m, not C.

Next, to calculate the magnitude of E in the mica, we can use the equation E = V/d, where V is the potential difference and d is the distance between the plates. However, in this case, we are not given the distance between the plates. Instead, we are given the plate area and capacitance of the parallel-plate capacitor. Therefore, we can use the equation E = [Q/A]/(k*Eo), where Q is the charge on the plates, A is the plate area, k is the dielectric constant, and Eo is the permittivity of free space. Plugging in the given values, we get E = [Q/0.013]/(5.4*8.85e-12) = 1.609 V/m. This is the correct value for E in the mica.

Moving on to the magnitude of the free charge on the plates, we can use the equation Q = CV, where C is the capacitance and V is the potential difference. Plugging in the given values, we get Q = (60e-12)*60 = 3.6e-9 C. Therefore, the magnitude of the free charge on the plates is 3.6 nanoCoulombs.

Finally, to calculate the magnitude of the induced surface charge on the mica, we can use the equation Q = k*Eo*E*A, where Q is the induced charge, k is the dielectric constant, Eo is the permittivity of free space, E is the electric field, and A is the plate area. Plugging in the given values, we get Q = 5.4*8.85e-12*1.609*0.013 = 6.244e-13 C. Therefore, the magnitude of the induced surface charge on the mica is 624.4 picoCoulombs.

In conclusion, to accurately