In the figure shown,after the switch $S$ is turned from position $A$ to position $B$,the energy dissipated in the circuit in terms of capacitance $C$ and total charge $Q$ is

  • A
    $\frac{1}{8}\frac{Q^2}{C}$
  • B
    $\frac{3}{8}\frac{Q^2}{C}$
  • C
    $\frac{5}{8}\frac{Q^2}{C}$
  • D
    $\frac{3}{4}\frac{Q^2}{C}$

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Similar Questions

In the given circuit,the charge $Q_2$ on the $2 \ \mu F$ capacitor changes as $C$ is varied from $1 \ \mu F$ to $3 \ \mu F$. $Q_2$ as a function of '$C$' is given properly by:

$n$ small drops of the same size are charged to $V$ volt each. If they coalesce to form a single large drop,then its potential will be:

What is the total electrostatic potential energy of the given system in $J$? (Given: $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9} \ N \cdot m^2/C^2$)

$A$ series combination of $N_1$ capacitors (each of capacity $C_1$) is charged to a potential difference $3V$. Another parallel combination of $N_2$ capacitors (each of capacity $C_2$) is charged to a potential difference $V$. The total energy stored in both combinations is the same. The value of $C_1$ in terms of $C_2$ is:

$(a)$ Determine the electrostatic potential energy of a system consisting of two charges $7 \; \mu C$ and $-2 \; \mu C$ (with no external field) placed at $(-9 \; cm, 0, 0)$ and $(9 \; cm, 0, 0)$ respectively.
$(b)$ How much work is required to separate the two charges infinitely away from each other?
$(c)$ Suppose that the same system of charges is now placed in an external electric field $E = A(1/r^2)$; $A = 9 \times 10^5 \; C \cdot m^{-2}$. What would the electrostatic energy of the configuration be?

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