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 can be expressed as :

Question

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 can be expressed as : (A) qb=qf(1-(1/√k)) (B) qb=qf(1-(1/k)) (C) qb=qf(1+(1/√k)) (D) qb=qf(1+(1/k))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/caacaf917f5e59ad1842817b0972da78-.png" /> When a dielectric is inserted in a capacitor Due to free charge vecE= vecE0 only After dielectric E'=(E0/k) qB=qf(1-(1/k))

Q. 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 $q_{b}$ can be expressed as :

$q_{b} = q_{f} (1 - \frac{1}{k})$

$q_{b} = q_{f} (1 - \frac{1}{k})$

$q_{b} = q_{f} (1 + \frac{1}{k})$

$q_{b} = q_{f} (1 + \frac{1}{k})$

When a dielectric is inserted in a capacitor Due to free charge
$E = E_{0}$ only After dielectric $E^{'} = \frac{E _{0}}{k}$
$q_{B} = q_{f} (1 - \frac{1}{k})$

Q. 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​ can be expressed as :

qb​=qf​(1−k​1​)

qb​=qf​(1−k1​)

qb​=qf​(1+k​1​)

qb​=qf​(1+k1​)

When a dielectric is inserted in a capacitor Due to free charge E=E0​ only After dielectric E′=kE0​​ qB​=qf​(1−k1​)

Q. 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 $q_{b}$ can be expressed as :

$q_{b} = q_{f} (1 - \frac{1}{k})$

$q_{b} = q_{f} (1 - \frac{1}{k})$

$q_{b} = q_{f} (1 + \frac{1}{k})$

$q_{b} = q_{f} (1 + \frac{1}{k})$

When a dielectric is inserted in a capacitor Due to free charge
$E = E_{0}$ only After dielectric $E^{'} = \frac{E _{0}}{k}$
$q_{B} = q_{f} (1 - \frac{1}{k})$