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What is the work needed to insert a dielectric with dielectric constant `kappa=2` into a parallel plate capacitor with capacitance `C` that is...


The mechanical work done on this system is going to be equal to the change in energy of the system.


The energy stored in the electric field of the capacitor is:


`E=1/2 CV^2`


A dielectric also increase the capacitance by a the factor `kappa` . Therefore the new capacitance `C_f` in relation to the original capacitance C would be:


`C_f=kappa*C`


`W=Delta E=E_f-E_i=1/2V^2(C_f-C)=1/2V^2(kappa*C-C)`


`W=1/2V^2(kappa-1)C`


`W=1/2V^2(2-1)C`


`W=1/2V^2C`


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