If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to
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- 1In the circuit shown, initially there is no charge on capacitors and keys $S _1$ and $S _2$ are open. The values of the capacitors are $C _1=10 \mu F , C _2=30 \mu F$ and $C _3= C _4=80 \mu F$.View Solution
Which of the statement($s$) is/are correct ?
$(1)$ The keys $S _1$ is kept closed for long time such that capacitors are fully charged. Now key $S _2$ is closed, at this time, the instantaneous current across $30 \Omega$ resistor (between points $P$ and $Q$ ) will be $0.2 A$
$(2)$ If key $S _1$ is kept closed for long time such that capacitors are fully charged, the voltage difference between points $P$ and $Q$ will be $10 V$.
$(3)$ At time $t =0$, the key $S _1$ is closed, the instantaneous current in the closed circuit will be $25 mA$.
$(4)$ If key $S _1$ is kept closed for long time such that capacitors are fully charged, the voltage across the capacitors $C _1$ will be $4 V$.
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- 10A parallel plate capacitor with plate area $'A'$ and distance of separation $'d'$ is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :View Solution
$\varepsilon(x)=\varepsilon_{0}+k x, \text { for }\left(0\,<\,x \leq \frac{d}{2}\right)$
$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for $\left(\frac{d}{2} \leq x \leq d\right)$