Q.A 6 μF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Energy stored in a capacitor U = ½CV² = 0.5 J.

The energy stored in a charged capacitor is

$U = \tfrac{1}{2}CV^{2}$

Substitute C = 100 μF = 100×10⁻⁶ F and V = 100 V: …

Charging a capacitor does work against the electric field building up between its plates; this work is stored as electrostatic potential energy.

When a capacitor of capacitance $C$ is charged to potential difference $V$ (charge $Q=CV$), external work is done to move charge against the growing field between the plates. This work does not appear as kinetic, heat, or magnetic e …

(a) 0.12 J …

Energy of a charged conductor $E = \tfrac{1}{2}CV^2$.

Charging a conductor to potential $V$ requires work, which is stored as electrostatic potential energy. Integrating the work to add successive charges $dq$ at potential $q/C$ gives

$E = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}.$

Using $Q = CV$, this can be written equivalently as …

Energy of a charged conductor is U = ½ CV² (equivalently ½ QV = Q²/2C).

A conductor of capacitance $C$ raised to potential $V$ holds charge $Q = CV$. To charge it, work is done against the growing potential. When the charge is $q$ the potential is $q/C$, so bringing an extra $dq$ costs $dW = (q/C)\,dq$.

Total work stored as electrostatic potential energy:

$U = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}.$

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The electrostatic energy stored in a charged capacitor is $U=\tfrac12CV^2$; direct substitution gives $4.5\times10^{-6}$ J.

Given: $C = 900\ \text{pF} = 900\times10^{-12}\ \text{F}$, $V = 100\ \text{V}$.

Formula

$U = \frac{1}{2}CV^2$

Substitution

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The electrostatic energy stored per unit volume of a region with field $E$ (in free space) is $u=\tfrac12 \varepsilon_0 E^2$.

This follows from the energy stored in a parallel-plate capacitor, $U=\tfrac12 C V^2$, divided by the volume $A d$ between the plates, using $C=\varepsilon_0 A/d$ and $E=V/d$:

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Use $U = \tfrac12 CV^2$ directly with $C = 700$ pF and $V = 50$ V.

Step 1 — Formula for energy stored in a charged capacitor

$U = \frac12 CV^2$

Step 2 — Substitute values

$C = 700\ \text{pF} = 700\times10^{-12}\ \text{F}, \qquad V = 50\ \text{V}$

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Energy stored in a capacitor is $U = Q^2/(2C)$, so $U \propto Q^2$.

The energy stored in a charged capacitor of capacitance $C$ carrying charge $Q$ is

$U = \frac{Q^2}{2C}$

For a fixed capacitor ($C$ constant), $U \propto Q^2$. If the charge is doubled, $Q_2 = 2Q_1$:

$U_2 = \frac{(2Q_1)^2}{2C} = \frac{4Q_1^2}{2C} = 4\left(\frac{Q_1^2}{2C}\right) = 4U_1$

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