Electrostatics Question 32
Question: Between the plates of a parallel plate condenser, a plate of thickness $ t _{1} $ and dielectric constant $ k _{1} $ is placed. In the rest of the space, there is another plate of thickness $ t _{2} $ and dielectric constant $ k _{2} $ . The potential difference across the condenser will be [MP PET 1993]
Options:
A) $ \frac{Q}{A{\varepsilon _{0}}}( \frac{t _{1}}{k _{1}}+\frac{t _{2}}{k _{2}} ) $
B) $ \frac{{\varepsilon _{0}}Q}{A}( \frac{t _{1}}{k _{1}}+\frac{t _{2}}{k _{2}} ) $
C) $ \frac{Q}{A{\varepsilon _{0}}}( \frac{k _{1}}{t _{1}}+\frac{k _{2}}{t _{2}} ) $
D) $ \frac{{\varepsilon _{0}}Q}{A}(k _{1}t _{1}+k _{2}t _{2}) $
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Answer:
Correct Answer: A
Solution:
Potential difference across the condenser
$ V=V _{1}+V _{2}=E _{1}t _{1}+E _{2}t _{2}=\frac{\sigma }{K _{1}{\varepsilon _{0}}}t _{1}+\frac{\sigma }{K _{2}{\varepsilon _{0}}}t _{2} $
therefore $ V=\frac{\sigma }{{\varepsilon _{0}}}( \frac{t _{1}}{K _{1}}+\frac{t _{2}}{K _{2}} )=\frac{Q}{A{\varepsilon _{0}}}( \frac{t _{1}}{K _{1}}+\frac{t _{2}}{K _{2}} ) $
