Q

# Help me please, - Electrostatics - JEE Main-6

Plate separation of a $15\mu F$  capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between the plates. Then new capacitance is given by

• Option 1)

$15\mu F$

• Option 2)

$20\mu F$

• Option 3)

$30\mu F$

• Option 4)

$25\mu F$

89 Views

As we have learned

If a number of dielectric slab inserted between the Plate -

$\dpi{100} {C}'=\frac{\epsilon _{0}A}{d-\left ( t_{1}+t_{2} +\cdots \right )+\left ( \frac{R_{1}}{k_{1}}+\frac{R_{2}}{k_{2}}+\cdots \right )}$

- wherein

given $C = \frac{\varepsilon _0A}{d}..........(1)$

Then by using

$C' = \frac{\varepsilon _0A}{d-t+\frac{t}{k}}= \frac{\varepsilon _0A}{2*10^{-3}-10^{-3}+\frac{10^{-3}}{2}}= 2/3*\varepsilon _0A*10^{3}$

; From equation (i) $C' =20\mu F$

Option 1)

$15\mu F$

Option 2)

$20\mu F$

Option 3)

$30\mu F$

Option 4)

$25\mu F$

Exams
Articles
Questions