A capacitance of a parallel plate air capacitor is \(10\mu\text{F}\). Dielectric constant of the medium to be introduced in between its plates is to double its capacitance is
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- \(2\)
- \(3\)
- \(4\)
- \(2.5\)
- \(1.5\)
The Correct Option is A
Solution and Explanation
Top Questions on Capacitors and Capacitance
- Energy stored in a capacitor is given by the equation \[ E = \frac{1}{2} C V^2 \] where: - \( C \) is the capacitance, - \( V \) is the voltage, - \( E \) is the energy stored. Given the values of \( C \), \( V \), and \( E \), determine the energy stored.}
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- Two capacitors $ C_1 = 4\mu F $ and $ C_2 = 6\mu F $ are connected in series across a 60 V battery. The potential difference across $ C_2 $ is:
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A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.
As soon as the circuit is completed by closing key S₁ (keeping S₂ open) charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference Vc (= q/C) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = VC). During this process of charging, the charge q on the capacitor changes with time t as
\(q = Q[1 - e^{-t/RC}]\)
The charging current can be obtained by differentiating it and using
\(\frac{d}{dx} (e^{mx}) = me^{mx}\)
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