$C _{1}=\frac{ A \varepsilon_{0}}{ d / 2}=\frac{2 A \varepsilon_{0}}{ d }= C$
$C _{2}=\frac{ KA \varepsilon_{0}}{ d / 2}=\frac{2 KA \varepsilon_{0}}{ d }=\frac{6 A \varepsilon_{0}}{ d }=3 C$
$C_{1}$ and $C_{2}$ are in series
$C _{\text {new }}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}=\frac{ C \times 3 C }{ C +3 C }=\frac{3 C }{4}$
$=\frac{3}{4} \times \frac{2 A \varepsilon_{0}}{d}=\frac{3}{2} \times \frac{A \varepsilon_{0}}{d}$
$C _{\text {new }}=\frac{3}{2} C _{\text {original }}$
$=\frac{3}{2} \times 4=6\, \mu\, F$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(a)$ When capacitor is air filled.
$(b)$ When capacitor is mica filled.
Current through resistor is $i$ and voltage across capacitor is $V$ then
| Column $I$ | Column $II$ |
| $(A)$ Point $P$ is situated midway between the wires. $Image$ | $(p)$ The magnetic fields $(B)$ at $P$ due to the currents in the wires are in the same direction. |
| $(B)$ Point $P$ is situated at the mid-point of the line joining the centers of the circular wires, which have same radii. $Image$ | $(q)$ The magnetic fields $(B)$ at $P$ due to the currents in the wires are in opposite directions. |
| $(C)$ Point $P$ is situated at the mid-point of the line joining the centers of the circular wires, which have same radii. $Image$ | $(r)$ There is no magnetic field at $P$. |
| $(D)$ Point $P$ is situated at the common center of the wires. $Image$ | $(s)$ The wires repel each other. |
