CBSE Class 12-science Answered

Difference between the properties with detailed explanation , derivation , figure and table for each like POTENTIAL DIFFERENCE between the plates of the parallel plate capacitor, ELECTRIC ENERGY stored between the plates of the parallel plate capacitor, CAPACITANCE of the capacitor , ELECTRIC FIELD between the plates of the parallel plate capacitor and CHARGE on the plates of the capacitor. in case of a DIELECTRIC of DIELECTRIC CONSTANT 'K' is introduced between the plates of a parallel plates capacitor and [i] battery remains CONNECTED across the CAPACITOR [ii] battery remains DISCONNECTED across the CAPACITOR .

Asked by pardeepkumar2281 | 18 May, 2018, 09:26: PM

Expert Answer

All the derivations you have asked in the questions are given in your NCERT Physics Text Book for class XII, Part-I ,

Chapter Two (Electrostatic Potential and Capacitance). To answer all the derivations you have asked simply I have to reproduce

the text book material. Instead, we request you to read the above mentioned chapters of your text book .

I will answer the part (1) battery remains connected across capacitor (2) battery remains disconnected across capacitor.

Figure shows the circuit to analyse the case of battery reamins connected/disconnected across capacitor. Let us assume we throw the switch to position a, so that the capacitor is getting charged. In time dt a charge dq = i×dt moves through any cross section of the circuit and is deposited on the positive plate of capacitor. The work E×dq done by the source of emf must equal the energy dissipated across Resistos i2Rdt and increase of energy dU in the capacitor. Energy U stored in the capacitor of capacitance C after charging of q Coulomb is given by U = q2/(2C).

Using conservation of energy we write, Edq = i2Rdt +d(q2 /(2C) ) = i2Rdt +(q/C)dq ................(1)

dividing (1) by dt, we get E(dq/dt) = i2R +(q/C) (dq/dt) ............................(2)

after substituting dq/dt = i in eqn.(2) and cancelling out common terms we get E = iR+(q/C) ................(3)

again substitte i = dq/dt in (3) , we get E = R(dq/dt) +(q/C) ..................(4)

we can rewrite eqn.(4) as

begin mathsize 12px style fraction numerator d q over denominator q minus E C end fraction space equals space minus fraction numerator d t over denominator R C end fraction.................... left parenthesis 5 right parenthesis end style

By integerating eqn.(5) with the initial condition q=0 at t=0, we get

begin mathsize 12px style q space equals space C E space left parenthesis space 1 space minus space e to the power of negative bevelled fraction numerator t over denominator R C end fraction end exponent right parenthesis space space space.................................. left parenthesis 6 right parenthesis end style

differentiating again (6) with respect to time t, we get

begin mathsize 12px style fraction numerator d q over denominator d t end fraction space equals space i space equals space E over R e to the power of negative bevelled fraction numerator t over denominator R C end fraction end exponent space space............................... left parenthesis 7 right parenthesis end style

Eqn.(6) gives the charge on the capacitor as a function of time. Eqn.(7) gives the current in the circuit as a function of time. The quantity RC in eqn.(6) and (7) has the dimension of time and is called as capacitive time constant τ_c of the circuit. ( τ_c = RC ) .

When the switch S in figure is closed on position b, the capacitor discharges through the resistor. Now there is no EMF in the circuit and Eqn(3) becomes

iR + (q/C) = 0 ................................(8)

eqn.(8) can be rewritten with the substitution i = dq/dt as R(dq/dt) + (q/C) = 0 .................(9)

eqn.(9) can be rearranged as dq/q = -dt/(RC) .............(10)

By integrating eqn.(10) with the initial condition charge at t=0 is q₀ = E×C , we get