If the above capacitor is connected across a 6.0 V battery, find (a) the charge supplied by the battery, (b) the induced charge on the dielectric and (c) the net charge appearing on one of the coated surfaces.

Open in App

Solution

Dielectric constant = 4, C

$= 1.42 μ F, v = 6 V$

Charge supplied,

$q = C v = 1.42 \times 10^{- 9} \times 6$

$= 8.52 \times 10^{- 9} C$

charge induced = $q (1 - \frac{1}{K})$

$= 8.52 \times 10^{- 9} \times (1 - 0.25)$

$= 6.39 \times 10^{- 9} = 6.4 n C$

Net charge appearing on one coated surface = 2.13 nC.

Suggest Corrections

Similar questions

Q. A capacitor of capacitance

5 μ F

is connected to a battery of emf

4 V .

Now this charged capacitor is disconnected with the battery and a dielectric of dielectric constant

2

is inserted between the plates. Find the magnitude of the charge induced on each surface of dielectric

Q. A capacitor of capacitance

100 μ F

is connected across a battery of emf

6.0 V

through a resistance of

20 k Ω

for

4 s

. The battery is then replaced by a thick wire. What will be the charge on the capacitor,

4 s

after the battery is disconnected?

Q. A capacitor stores 50 µC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100 µC flows through the battery. Find the dielectric constant of the material inserted.

Q. An air capacitor is connected to a battery of emf V. A dielectric medium of dielectric constant K is introduced between the plates of this capacitor, with the battery still connected. The final charge on the capacitor plates relative to the previous charge

Two capacitors of $0.5 μ F$ and $1 μ F$ are connected in parallel across a battery. If the charge on $0.5 μ F$ is $50 μ C$ , the charge on the other capacitor is :

Join BYJU'S Learning Program

If the above capacitor is connected across a 6.0 V battery, find (a) the charge supplied by the battery, (b) the induced charge on the dielectric and (c) the net charge appearing on one of the coated surfaces.

Dielectric constant = 4, C =1.42 μF, v=6V Charge supplied, q=Cv=1.42×10−9×6 =8.52×10−9C charge induced = q(1−1K) =8.52×10−9×(1−0.25) =6.39×10−9=6.4 nC Net charge appearing on one coated surface = 2.13 nC.