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A capacitor $P$ with capacitance $10 \times 10^{-6} \text{ F}$ is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \text{ F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} \text{ C}$ (assume capacitor $Q$ does not have any charge initially), the value of $\alpha$ is _________.

Capacitor Charge Calculation

This problem involves calculating the charge redistribution between two capacitors connected in parallel after one is initially charged.

Initial Charge Calculation

  • Capacitor P has capacitance $C_P = 10 \times 10^{-6} \text{ F}$ and initial potential difference $V_P = 6.0 \text{ V}$.
  • The initial charge on capacitor P is calculated using the formula $Q = CV$.
  • $Q_P = C_P \times V_P = (10 \times 10^{-6} \text{ F}) \times (6.0 \text{ V}) = 60 \times 10^{-6} \text{ C}$.
  • Capacitor Q has capacitance $C_Q = 20 \times 10^{-6} \text{ F}$ and is initially uncharged ($Q_Q = 0$).

Charge Redistribution and Equilibrium

When the charged capacitor P is connected across capacitor Q, they form a parallel circuit. Charge flows from P to Q until they reach a common final potential difference, $V_f$. Charge is conserved during this process.

  • Total initial charge = Charge on P + Charge on Q = $60 \times 10^{-6} \text{ C} + 0 = 60 \times 10^{-6} \text{ C}$.
  • The equivalent capacitance in parallel is $C_{eq} = C_P + C_Q = (10 \times 10^{-6}) + (20 \times 10^{-6}) = 30 \times 10^{-6} \text{ F}$.
  • The final common potential difference $V_f$ is found using $V_f = \frac{\text{Total Charge}}{C_{eq}}$.
  • $V_f = \frac{60 \times 10^{-6} \text{ C}}{30 \times 10^{-6} \text{ F}} = 2.0 \text{ V}$.

Final Charge on Capacitor Q

The charge on capacitor Q at equilibrium ($Q_{Q,f}$) is calculated using its capacitance and the final common potential:

  • $Q_{Q,f} = C_Q \times V_f = (20 \times 10^{-6} \text{ F}) \times (2.0 \text{ V}) = 40 \times 10^{-6} \text{ C}$.
  • The question states this charge is $\alpha \times 10^{-5} \text{ C}$.
  • Therefore, $40 \times 10^{-6} \text{ C} = \alpha \times 10^{-5} \text{ C}$.
  • Rewriting $40 \times 10^{-6}$ as $4 \times 10^{-5}$, we get $4 \times 10^{-5} \text{ C} = \alpha \times 10^{-5} \text{ C}$.
  • Comparing both sides gives $\alpha = 4$.
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Important Questions from Electricity and Magnetism

  1. The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by,
    $E_y = 20 \sin(3 \times 10^6 x - 4.5 \times 10^{14} t) \text{ V/m}$
    (where x, t and other values have S.I. units). The dielectric constant of the medium is _________.
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  3. A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by $30^\circ$ from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is _________ mm.
    $\left(\text{Take } \pi = \frac{22}{7}\right)$
  4. A cylindrical conductor of length 2 m and area of cross-section $0.2 \text{ mm}^2$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \text{ m}^2\text{/V.s}$. The value of $\alpha$ is :
    (electron concentration $= 5 \times 10^{28} \text{/m}^3$ and electron charge = $1.6 \times 10^{-19} \text{ C}$)
  5. There are three co-centric conducting spherical shells $A$, $B$ and $C$ of radii $a$, $b$ and $c$ respectively ($c > b > a$) and they are charged with charge $q_1$, $q_2$ and $q_3$ respectively. The potentials of the spheres $A$, $B$ and $C$ respectively, are :
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