A thin metal plate $P$ is inserted half way between the plates of a parallel plate capacitor of capacitance $C$ in such a way that it is parallel to the two plates. The capacitance now becomes
Easy
Download our app for free and get started
(a) Thin metal plates doesn't affect the capacitance.
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Six point charges are placed at the vertices of a regular hexagon of side $a$ as shown. If $E$ represents electric field and $V$ represents electric potential at $O$, then
A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $0$ to $3d$
Consider a parallel plate capacitor of $10\,\mu \,F$ (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant $4$, as shown in the figure. The capacity of the capacitor changes to.......$\mu \,F$
A bullet of mass $2\, gm$ is having a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$
In the circuit shown here $C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu F$)finally on $C_2$ ?