1
JEE Main 2016 (Offline)
+4
-1
A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge $$Q$$ (having a charge equal to the sum of the charges on the $$4$$ $$\mu \,F$$ and $$9$$ $$\mu \,F$$ capacitors), at a point distance $$30$$ $$m$$ from it, would equal :

A
$$420N/C$$
B
$$480N/C$$
C
$$240N/C$$
D
$$360N/C$$
2
JEE Main 2015 (Offline)
+4
-1
In the given circuit, charges $${Q_2}$$ on the $$2\mu F$$ capacitor changes as $$C$$ is varied from $$1\,\mu F$$ to $$3\mu F.$$ $${Q_2}$$ as a function of $$'C'$$ is given properly by:
$$\left( {figures\,\,are\,\,drawn\,\,schematically\,\,and\,\,are\,\,not\,\,to\,\,scale} \right)$$

A
B
C
D
3
JEE Main 2014 (Offline)
+4
-1
A parallel plate capacitor is made of two circular plates separated by a distance $$5$$ $$mm$$ and with a dielectric of dielectric constant $$2.2$$ between them. When the electric field in the dielectric is $$3 \times {10^4}\,V/m$$ the charge density of the positive plate will be close to:
A
$$6 \times {10^{ - 7}}\,\,C/{m^2}$$
B
$$3 \times {10^{ - 7}}\,\,C/{m^2}$$
C
$$3 \times {10^4}\,\,C/{m^2}$$
D
$$6 \times {10^4}\,\,C/{m^2}$$
4
JEE Main 2013 (Offline)
+4
-1
Two capacitors $${C_1}$$ and $${C_2}$$ are charged to $$120$$ $$V$$ and $$200$$ $$V$$ respectively. It is found that connecting them together the potential on each one can be made zero. Then
A
$$5{C_1} = 3{C_2}$$
B
$$3{C_1} = 5{C_2}$$
C
$$3{C_1} + 5{C_2} = 0$$
D
$$9{C_1} = 4{C_2}$$
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