How Do You Calculate Capacitance for Split Capacitors in PhET Simulations?

[physocs1]

Homework Statement

A= 400mm^2 d=7.5mm v=1.5 kappa= 4.0 offset=8.00mm [/B]
Click on the dielectric tab. Configure the single capacitor circuit by using the parameters given in the circuit diagram to the right. This capacitor with its offset dielectric slab is equivalent to two component capacitors arranged in parallel. Each of the component capacitors (right and left) is considered to be a capacitor for which the expression C = Aεoκ/d can be used to compute its contributing capacitance. For each of the components, you must use the dielectric constant of the material within its domain (either air or the material for which κ was set to the value 4.0). In table 2A on the results sheet you will need to determine the width and depth of each of the component capacitors’ effective plate areas. To do this, first find the common plate depth (remember that the full plate is a square, so the depth and full width should be the square root of the plate area). Then use the dielectric offset to find each of the component plate widths. Enter these quantities in table 2A. You do not need to show these calculations since they are fairly trivial. Then use the expression C = Aεoκ/d to compute the effective capacitance of each of the two capacitor components. Enter these quantities as CL and CR in table 2A, (rounded to 3 significant figures). Show your calculations in the spaces provided.

I'm not understanding what the lab instructions means by calculating the right side and left side of the capacitance. other than that I know how to do the lab.

Website: https://phet.colorado.edu/en/simulation/legacy/capacitor-lab


Thank you

 

[Student100]

Homework Statement

A= 400mm^2 d=7.5mm v=1.5 kappa= 4.0 offset=8.00mm [/B]
Click on the dielectric tab. Configure the single capacitor circuit by using the parameters given in the circuit diagram to the right. This capacitor with its offset dielectric slab is equivalent to two component capacitors arranged in parallel. Each of the component capacitors (right and left) is considered to be a capacitor for which the expression C = Aεoκ/d can be used to compute its contributing capacitance. For each of the components, you must use the dielectric constant of the material within its domain (either air or the material for which κ was set to the value 4.0). In table 2A on the results sheet you will need to determine the width and depth of each of the component capacitors’ effective plate areas. To do this, first find the common plate depth (remember that the full plate is a square, so the depth and full width should be the square root of the plate area). Then use the dielectric offset to find each of the component plate widths. Enter these quantities in table 2A. You do not need to show these calculations since they are fairly trivial. Then use the expression C = Aεoκ/d to compute the effective capacitance of each of the two capacitor components. Enter these quantities as CL and CR in table 2A, (rounded to 3 significant figures). Show your calculations in the spaces provided.

I'm not understanding what the lab instructions means by calculating the right side and left side of the capacitance. other than that I know how to do the lab.

Website: https://phet.colorado.edu/en/simulation/legacy/capacitor-lab


Thank you

Can't open the simulation on my phone.

It sounds as though you have a cap that's partially filled with a dielectric. You're given an outline of how they want you to calculate the contribution of each part treated as two caps in parallel.

Draw the picture of the whole system, label the parts, and see if you can see how they would be broken up. Then draw a new circuit diagram.

 

[andrevdh]

I get the impression that the single capacitor can be replaced by a parallel combination of two capacitors, one on the left and one on the right. You need to calculate the two capacitances, the one on the left and the one on the right in the parallel combination.

 

[physocs1]

Ok, I understand now. Thank-you all