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Electric field intensities in dielectric slab

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>The plates of a parallel plate capacitor are separated by a distance d = 1 cm. Two parallel sided dielectric slabs of thickness 0.7 cm and 0.3 cm fill the space between the plates. If the dielectric constants of the two slabs are 3 and 5 respectively and a potential difference of 440V is applied across the plates. Find : the electric field intensities in each of the slabs.

My try

Now how to proceed ?

asked Jan 2, 2017 in Physics Problems by koolman (4,286 points)
reshown May 16 by sammy gerbil
Please can you explain your calculation?

Assuming that your 2 equations are correct, from the 1st you can find $Q/\epsilon_0 A$ and then substitute into the 2nd equation.
In the first equation . I have just equated potentials . As electric field inside the capacitor is (Q/A$\epsilon$$_0$$\epsilon$$_r$)  and multiplying it with d we get potential .
The second equation is of electric field , but we have to find electric field intensities .
The 2nd "equation" is in fact 2 equations : $E_1=Q/\epsilon_1 \epsilon_0 A$ and similar for $E_2$.

Apart from this your method of solution is correct. Please post your answer for me to upvote it.

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