Solve

Guides

0

You visited us 0 times! Enjoying our articles? Unlock Full Access!

Question

In the circuit below, if a dielectric is inserted into $C_{2}$ then the charge on $C_{1}$ will

Increase
Decrease
Remain same
Be halved

A

Increase

B

Remain same

C

Be halved

D

Decrease

Open in App

Solution

Verified by Toppr

Initially wire between both capacitors is neutral, as the source is connected. Positive charge moves toward B and negative charge move toward A.

when the dielectric is placed between the capacitor plate charge increase, due to which wire ends B having more positive charge flow toward it and more positive charge flow toward B.

Hence, charge on capacitor $C_{1}$ also increase

Was this answer helpful?

6

Similar Questions

Q1

In the circuit below, if a dielectric is inserted into

C_{2}

then the charge on

C_{1}

will

View Solution

Q2

In given circuit,

C_{1} = C_{2} = C_{3} = C

initially. Now, a dielectric slab of dielectric constant

K = \frac{3}{2}

is inserted in

C_{2}

.
The equivalent capacitance become

View Solution

Q3

Two identical capacitor

C_{1}

and

C_{2}

are connected in series with a battery. They are fully charged. Now dielectric slab is inserted between the plates of

C_{2}

. The potential difference across

C_{1}

will

View Solution

Q4

Two identical capacitors

C_{1}

and

C_{2}

are connected in series with a battery as shown in the figure. They are fully charged. Now dielectric slab is inserted between the plates of

C_{2}

. The potential difference across

C_{1}

will now

View Solution

Q5

In the circuit shown in figure, if

C_{1} = C_{2} = 2 μ F

,then charges present on

C_{1} and C_{2}

will be

View Solution