capacitanca of a spherical capacitor with dielectric between both the capacitor.dielectric is concentric to the spheres.dielectric is partially between both the sphere
Professor l want to correct something about the minus sign we use in the integral. Voltage decrease because it's proportional 1/R not because of the minus sign. That sign indicates that the force in the first place is the electrical force not the force we apply to push the charge near another charge. The electrical force acts always in the opposite way according to the displacement. So, from the scaler product between the electric field and the displacement, cosinus has an angle of 180 degree which indicates the negative sign. Your process goes truely but the point of view is incorrect. l just want to share and thank you for all your efforts to create this platform.
I know that the problem is corectly solved, but what i canot understand is that when a dielectric is placed on an electric field on the dielectric is created a ''trapped'' charge on its surface and on it's volume wich are often symbolized with the greek letters ''σ'' or ''ρ'' and they are connected with the polarization via the types σ=P η and ρ=-grad P. These charge densities do not contibute in the E ? Thank you
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capacitanca of a spherical capacitor with dielectric between both the capacitor.dielectric is concentric to the spheres.dielectric is partially between both the sphere
Professor l want to correct something about the minus sign we use in the integral. Voltage decrease because it's proportional 1/R not because of the minus sign. That sign indicates that the force in the first place is the electrical force not the force we apply to push the charge near another charge. The electrical force acts always in the opposite way according to the displacement. So, from the scaler product between the electric field and the displacement, cosinus has an angle of 180 degree which indicates the negative sign. Your process goes truely but the point of view is incorrect. l just want to share and thank you for all your efforts to create this platform.
Why E is this quantity? the trapped charge on dielectric does not produce a second field which must be added to E?
The electric field only exists between the 2 capacitor plates. (there is only one field)
I know that the problem is corectly solved, but what i canot understand is that when a dielectric is placed on an electric field on the dielectric is created a ''trapped'' charge on its surface and on it's volume wich are often symbolized with the greek letters ''σ'' or ''ρ'' and they are connected with the polarization via the types σ=P η and ρ=-grad P. These charge densities do not contibute in the E ? Thank you
The dielectric typically will set up and electric field that opposes the original field and will therefore reduce the strength of the field.
How is the thickness of the dielectric related to the capacitance of the system?
I think it doesn't depend upon its thickness , rather it only depends upon its dielectric constant
inversely proportional
Professor why can't we use Gauss's law for the electric field of the capacitor?
Because there is a dielectric material between two electrodes
İlker Aydın we can use gauss's law....we just have to replace 4π€ by k4π€....the answer will be the same....