Er1=3Er2=51d₁=2cmd₂=6cmFig 1. Parallel Plate Capacitor with two dielectric materials. The area of the capacitor is0.24m². The top plate (electrode) is positively charged.Question 1: Use Gauss law to find an expression for the electric field in each of the dielectricmaterials. Put the value of electric field in dielectric 2 if there is a charge of Q=2.3μC on the platesin your answer table.

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Fig 1. Parallel Plate Capacitor with two dielectric materials. The area of the capacitor is 0.24m². The top plate (electrode) is positively charged. in each of the dielectric Question 1: Use Gauss law to find an expression for the electric field materials. Put the value of electric field in dielectric 2 if there is a charge of Q=2.3 μC on the plates in your answer table.

Fig 1. Parallel Plate Capacitor with two dielectric materials. The area of the capacitor is 0.24m². The top plate (electrode) is positively charged. in each of the dielectric Question 1: Use Gauss law to find an expression for the electric field materials. Put the value of electric field in dielectric 2 if there is a charge of Q=2.3 μC on the plates in your answer table.

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Er1=3
Er2=5
1d₁=2cm
d₂=6cm
Fig 1. Parallel Plate Capacitor with two dielectric materials. The area of the capacitor is
0.24m². The top plate (electrode) is positively charged.
Question 1: Use Gauss law to find an expression for the electric field in each of the dielectric
materials. Put the value of electric field in dielectric 2 if there is a charge of Q=2.3μC on the plates
in your answer table.

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