Question 289: Electrostatics & Capacitance

Question

If q
f
is the free charge on the capacitor plates and q
b
is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge q
b
an be expressed as :

Accepted Answer

When a dielectric is inserted in a capacitor
Due to free charge $$\overrightarrow E  = {\overrightarrow E _0}$$ only
After dielectric $$E' = {{{E_0}} \over k}$$
$${q_B} = {q_f}\left( {1 - {1 \over k}} \right)$$

Answer

$${q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)$$

Answer

$${q_b} = {q_f}\left( {1 - {1 \over k}} \right)$$

Answer

$${q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)$$

Answer

$${q_b} = {q_f}\left( {1 + {1 \over k}} \right)$$

Electrostatics & Capacitance

If q f is the free charge on the capacitor plates and q b is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge q b an be expressed as :

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