After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
Medium
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Battery is removed, charge on plate is constant.By placing a dielectric slab between the plates,capacitance increase so the potential difference between the plates and energy stored will decreases.
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The plates of a parallel plate capacitor of capacity $50\,\mu C$ are charged to a potential of $100\;volts$ and then separated from each other so that the distance between them is doubled. How much is the energy spent in doing so
A capacitor of $4\, \mu F$ is connected to a $15\, V$ supply through $1$ mega ohm resistance. The time taken by the capacitor to charge upto $63.2\%$ of its final charge will be......$s$
A charge $ + q$ is fixed at each of the points $x = {x_0},\,x = 3{x_0},\,x = 5{x_0}$..... $\infty$, on the $x - $axis and a charge $ - q$ is fixed at each of the points $x = 2{x_0},\,x = 4{x_0},x = 6{x_0}$,..... $\infty$. Here ${x_0}$ is a positive constant. Take the electric potential at a point due to a charge $Q$ at a distance $r$ from it to be $Q/(4\pi {\varepsilon _0}r)$. Then, the potential at the origin due to the above system of charges is