DC Pandey Solutions: Capacitors | Physics Class 12 - NEET PDF Download

Introductory Exercise 22.1

Q.1. Find the dimensions of capacitance.

Sol.

= [M^-1L^-2T⁴A²]

Q.2. No charge will flow when two conductors having the same charge are connected to each other. Is this statement true or false?

Sol. Charge does not flow if their potentials are same.

Q.3. Two metallic plates are kept parallel to one another and charges are given to them as shown in the figure. Find the charge on all four faces.

Sol. Charge on outermost surfaces

Hence, charges are as shown below

Q.4. Charges 2q and -3g are given to two identical metal plates of area of cross-section A. The distance between the plates is d. Find the capacitance and potential difference between the plates.

Sol. Charge on outermost surfaces.

Hence charge on different faces are as shown below.

Electric field and hence potential difference between the two plates is due to ±2.5 q

Introductory Exercise 22.2

Q.1. Find the charge stored in all the capacitors.

Sol: All three capacitors are in parallel wit h the battery. PD across each of them is 10V. So, apply q = CV for all of them.

Q.2. Find the charge stored in the capacitor.

Sol. Capacitor and resistor both are in parallel with the battery. PD across the capacitor is 10V. Now apply q = CV.

Q.3. Find the charge stored in the capacitor.

Sol. In steady state current flows in lower loop of the circuit.

Now, potential difference across capacitor = potential difference across 4Ω resistance.
= iR
= (3)(4)=12V
q = CV = (2μF) (12 V)
= 24μC

Q.4. A 2μF capacitor and a 2μF capacitor are connected in series across a 1200 V supply line.
(a) Find the charge on each capacitor and the voltage across them.
(b) The charged capacitors are disconnected from the line and from each other and reconnected with terminals of like sign together. Find the final charge on each and the voltage across them.

Sol. 

= 800μC
In series, q remains same
q₁ = q₂ = 800 μC

and

(b) Now total charge will become 1600 μC. This will now distribute in direct ratio of capacity

They will have a common potential (in parallel) given by.

Q.5. A 100 μF capacitor is charged to 100 V. After the charging, the battery is disconnected. The capacitor is then connected in parallel to another capacitor. The final voltage is 20 V. Calculate the capacity of the second capacitor.

Sol. Charge, q = CV = 10⁴μC
In parallel common potential is given by

Solving this equation we get
C = 400μF

Introductory Exercise 22.3

Q.1. An uncharged capacitor C is connected to a battery through a resistance R. Show that by the time the capacitor gets fully charged, the energy dissipated in R is the same as the energy stored in C.

Sol. Charge supplied by the battery,
q= CV
Energy supplied by the battery,
E = qV =CV²
Energy stored in the capacitor,

∴ Energy dissipated across R in the form of heat = E - U

Q.2. How many time constants will elapse before the current in a charging RC circuit drops to half of its initial value?

Sol.

Q.3. A capacitor of capacitance C is given a charge q₀. At time t = 0 it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the first capacitor and the second capacitor as a function of time f. Also plot the corresponding q-t graphs.

Sol. Both capacitors have equal capacitance. Hence half-half charge distribute over both the capacitors.

q₁ decreases exponentially from
While q₂ increases exponentially from  Corresponding graphs and equation are given in the answer. Time constant of two exponential equations will be

Q.4. A capacitor of capacitance C is given a charge q₀. At time f = 0, it is connected to a battery of emf E through a resistance R. Find the charge on the capacitor at time t.

Sol. q_i = q₀ q_f = EC
Now charge on capacitor changes from q_i to q_f exponentially.

Q.5. Determine the current through the battery in the circuit shown in figure

(a) immediately after the switch S is closed
(b) after a long time.

Sol. 

(a) Immediately after the switch is closed whole current passes through C₁ 
i = E/R₁
(b) Long after switch is closed no current will pass through C₁ and C₂.

Q.6. For the circuit shown in the figure find

(a) the initial current through each resistor
(b) steady-state current through each resistor
(c) final energy stored in the capacitor
(d) the time constant of the circuit when the switch is opened.

Sol. 

(a) At t = 0 equivalent resistance of capacitor is zero. R₁ and R₂ are in parallel across the battery PD across each is E.

(b) In steady state no current flow through capacitor wire. PD across R₁ is E.

(c) In steady state potential difference across capacitor is E.
∴
(d) When switch is opened, capacitor is discharged through resistors R₁ and R₂