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racecar_

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- Homework Statement
- Two parallel plates, having a vacuum in between them, are separated by d=0.01 m apart and are connected to a battery maintaining 9V between the plates and a charge of Q0 magnitude on each plate. The separation between the plates is then raised to 2d, the battery voltage is increased to 18V, and a dielectric is added with dielectric constant 2. What is the energy density difference after the addition of the dielectric?

- Relevant Equations
- energy density = (1/2)εE^2

I am confused about how the electric field changes in this problem - is E' = E/K

When I solve it this way, the answer is incorrect:

change in energy density = (1/2)ε(E'

What am I doing wrong? Thanks.

_{e}=E/2? Is E = V/d a correct usage?When I solve it this way, the answer is incorrect:

change in energy density = (1/2)ε(E'

^{2}- E^{2}) = (1/2)ε(E^{2}/4 - E^{2}) = (1/2)ε(-3/4)(V/2d)^{2}.What am I doing wrong? Thanks.