1
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1 Two identical thin metal plates has charge $$q_{1}$$ and $$q_{2}$$ respectively such that $$q_{1}>q_{2}$$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

A
$$\frac{\left(q_{1}+q_{2}\right)}{C}$$
B
$$\frac{\left(q_{1}-q_{2}\right)}{C}$$
C
$$\frac{\left(q_{1}-q_{2}\right)}{2 C}$$
D
$$\frac{2\left(q_{1}-q_{2}\right)}{C}$$
2
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1 A slab of dielectric constant $$\mathrm{K}$$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $$\frac{3}{4} \mathrm{~d}$$, where $$\mathrm{d}$$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :

(Given $$\mathrm{C}_{0}$$ = capacitance of capacitor with air as medium between plates.)

A
$$\frac{4 K C_{0}}{3+K}$$
B
$$\frac{3 K C_{0}}{3+K}$$
C
$$\frac{3+K}{4 K C_{0}}$$
D
$$\frac{K}{4+K}$$
3
JEE Main 2022 (Online) 28th July Morning Shift
+4
-1 Two capacitors, each having capacitance $$40 \,\mu \mathrm{F}$$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $$\mathrm{K}$$ such that the equivalence capacitance of the system became $$24 \,\mu \mathrm{F}$$. The value of $$\mathrm{K}$$ will be :

A
1.5
B
2.5
C
1.2
D
3
4
JEE Main 2022 (Online) 26th July Evening Shift
+4
-1 A source of potential difference $$V$$ is connected to the combination of two identical capacitors as shown in the figure. When key '$$K$$' is closed, the total energy stored across the combination is $$E_{1}$$. Now key '$$K$$' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $$E_{2}$$. The ratio $$E_{1} / E_{2}$$ will be : A
$$\frac{1}{10}$$
B
$$\frac{2}{5}$$
C
$$\frac{5}{13}$$
D
$$\frac{5}{26}$$
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