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 Between the plates of a parallel plate condenser, a plate of thickness  t_1and dielectric constant k_1 is placed. In the rest of the space, there is another plate of thickness  t_2and dielectric constant k_2.  The potential difference across the condenser will be

  • Option 1)

    (Q/A\varepsilon _0)(\frac{t_1}{k_1}+\frac{t_2}{k_2})

  • Option 2)

    (Q\varepsilon _0/A)(\frac{t_1}{k_1}+\frac{t_2}{k_2})

  • Option 3)

    (Q/\varepsilon _0A)(\frac{k_1}{t_1}+\frac{k_2}{t_2})

  • Option 4)

    (Q\varepsilon _0/A)({t_1}{k_1}+{t_2}{k_2})

 

Answers (1)

best_answer

As we have learned

If a number of dielectric slab inserted between the Plate -

{C}'=\frac{\epsilon _{0}A}{d-\left ( t_{1}+t_{2} +\cdots \right )+\left ( \frac{R_{1}}{k_{1}}+\frac{R_{2}}{k_{2}}+\cdots \right )}

- wherein

 

 

         Potential difference across the condenser

V=V_1+V_2=E_1t_+E_2t_2= \frac{\sigma }{k_1\varepsilon _0}t_1+\frac{\sigma }{k_2\varepsilon _0}t_2= \frac{Q}{A\varepsilon _0}\left ( \frac{t_1}{k_1}+\frac{t_2}{k_2} \right ) 


Option 1)

(Q/A\varepsilon _0)(\frac{t_1}{k_1}+\frac{t_2}{k_2})

Option 2)

(Q\varepsilon _0/A)(\frac{t_1}{k_1}+\frac{t_2}{k_2})

Option 3)

(Q/\varepsilon _0A)(\frac{k_1}{t_1}+\frac{k_2}{t_2})

Option 4)

(Q\varepsilon _0/A)({t_1}{k_1}+{t_2}{k_2})

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