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Physics · JEE | Electrostatic Potential and Capacitance

हिन्दी

An uncharged capacitor of capacitance \( C \) is connected in series with a resistor and a DC source of potential \( V_0 \). What is the total work done by the battery during the charging process of the capacitor?

  1. A
    \( \frac{1}{2} C V_0^2 \)
  2. B
    \( \frac{1}{4} C V_0^2 \)
  3. C
    \( 2 C V_0^2 \)
  4. D
    \( C V_0^2 \)
View Answer & Solution

Correct answer · Option D

\( C V_0^2 \)

Explanation

The total charge supplied by the battery is \( Q = CV_0 \). The work done by the battery source is \( W_{battery} = QV_0 = (CV_0)V_0 = CV_0^2 \). (Note: only half is stored as potential energy, the rest is dissipated as heat).

Related questions

A parallel-plate capacitor with vacuum between the plates has a capacitance \( C_0 \). A dielectric slab of dielectric constant \( K = 5 \) and thickness equal to \( \frac{3}{4}d \) is inserted between the plates, where \( d \) is the separation between the plates. What is the new capacitance \( C \) of the capacitor?→What is the energy density (energy stored per unit volume) in the electric field of a parallel plate capacitor in vacuum, given that the electric field strength between the plates is \( 3 \times 10^4\text{ V/m} \)? (Take \( \varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2} \))→What is the potential energy \( U(\theta) \) of an electric dipole of dipole moment \( \mathbf{p} \) in a uniform electric field \( \mathbf{E} \), taking zero potential energy at \( \theta = \pi/2 \)?→A linear isotropic dielectric slab is placed in an external uniform electric field \( E_0 \). If the electric susceptibility of the dielectric is \( \chi_e = 4 \), and the magnitude of the polarization vector is \( P = 1.2 \times 10^{-7}\text{ C/m}^2 \), what is the magnitude of the external electric field \( E_0 \) in \( \text{V/m} \)? (Take \( \varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2} \))→
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