JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 8)
If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as :
$${q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)$$
$${q_b} = {q_f}\left( {1 - {1 \over k}} \right)$$
$${q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)$$
$${q_b} = {q_f}\left( {1 + {1 \over k}} \right)$$
Explanation
_25th_July_Evening_Shift_en_8_2.png)
When a dielectric is inserted in a capacitor
Due to free charge $$\overrightarrow E = {\overrightarrow E _0}$$ only
After dielectric $$E' = {{{E_0}} \over k}$$
$${q_B} = {q_f}\left( {1 - {1 \over k}} \right)$$
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