165713
In the given electrical circuit, if the switch $S$ is closed then the maximum energy stored in the inductors is:
1 $3 \mathrm{~J}$
2 $9 \mathrm{~J}$
3 $12 \mathrm{~J}$
4 $6 \mathrm{~J}$
Explanation:
: Energy in First capacitor, $\mathrm{E}_{1}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ $\mathrm{E}_{1}=\frac{1}{2} \times \frac{4^{2}}{1}=8 \mathrm{~J}$ Energy in second capacitor, $E_{2}=\frac{1}{2} \cdot \frac{Q^{2}}{C}$ $=\frac{1}{2} \cdot \frac{2}{2}^{2}$ $\mathrm{E}_{2} =1 \mathrm{~J}$ Total energy, $\mathrm{E}_{\mathrm{o}}=\mathrm{E}_{1}+\mathrm{E}_{2}$ $=8+1$ $=9 \mathrm{~J}$ Now switch is closed then the common potential of capacitors $\mathrm{V}_{\text {common }}= \frac{\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}_{1}}{\mathrm{C}_{1}}=\frac{4}{1}=4 \mathrm{~V}$ $\mathrm{~V}_{2}=\frac{\mathrm{Q}_{2}}{\mathrm{C}_{2}}=\frac{2}{2}=1 \mathrm{~V}$ $\mathrm{~V}_{\text {common }}=\frac{1 \times 4+2 \times 1}{1+2}$ $=2 \mathrm{~V}$ Hence, the new arrangement of energy, $\mathrm{E}_{1}=\frac{1}{2} \mathrm{C}_{1} \mathrm{~V}_{\text {Common }}^{2}=\frac{1}{2} \times 1 \times 4=2 \mathrm{~J}$ $\mathrm{E}_{2}=\frac{1}{2} \mathrm{C}_{2} \mathrm{~V}_{\text {common }}^{2}=\frac{1}{2} \times 2 \times 4=4 \mathrm{~J}$ Now from conservation of energy stored in the inductor $\mathrm{E}_{1} \square \mathrm{E}_{\mathrm{o}}-\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right)$ $=9-(2+4)$ $=3 \mathrm{~J}$
[AP EAMCET (21.04.2019) Shift-I]
Capacitance
165714
In an oscillating LC circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is
1 $\frac{Q}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{3}}$
3 Q
4 $\frac{\mathrm{Q}}{\sqrt{2}}$
Explanation:
: The electric energy stored when energy is equally distributed. $\mathrm{U}_{\mathrm{E}}^{\prime}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}$ The energy stored in the capacitor when it is maximum charged. $\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ At any time electrostatic energy is equal to magnetic energy. $=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{c}}+\frac{1}{2} \mathrm{LI}^{2}$ $\because \frac{\mathrm{q}^{2}}{2 \mathrm{c}}=\frac{\mathrm{LI}^{2}}{2}$ According to question - $\mathrm{U}_{\mathrm{E}}=2 \mathrm{U}_{\mathrm{E}}^{\prime}$ $\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=2 \times \frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}$ $\frac{\mathrm{Q}^{2}}{2}=\mathrm{q}^{2}$ $\mathrm{q}=\sqrt{\frac{\mathrm{Q}^{2}}{2}}$ $\mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}$
[AP EAMCET (20.04.2019) Shift-1]
Capacitance
165715
The total electrostatic energy stored in both the capacitor is
1 $9 \mu \mathrm{J}$
2 $40.5 \mu \mathrm{J}$
3 $13.5 \mu \mathrm{J}$
4 $18 \mu \mathrm{J}$
Explanation:
: Given, capacitance $\mathrm{C}_{1}=3 \mu \mathrm{F}, \mathrm{C}_{2}=6 \mu \mathrm{F}$ Capacitors are connected in series combination, So, equivalent capacitance $\left(\mathrm{C}_{\mathrm{eq}}\right)=\frac{\mathrm{C}_{1} \times \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ $\mathrm{C}_{\mathrm{eq}}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}$ Since, total electrostatic energy $(U)=\frac{1}{2} C_{\text {eq }} V^{2}$ $\mathrm{U}=\frac{1}{2} \times 2 \times(-3)^{2} \times 10^{-6}$ $\mathrm{U}=9 \mu \mathrm{J}$
[UPSEE - 2017]
Capacitance
165716
Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch $S$. If now the switch $S$ is opened and the space between the plates is filled with dielectric of relative permittivity $\varepsilon_{\mathrm{r}}$, then
1 the potential difference as well as charge on each capacitor goes up by a factor $\varepsilon_{\mathrm{r}}$
2 the potential difference as well as charge on each capacitor goes down by a factor $\varepsilon_{\mathrm{r}}$
3 the potential difference across a remains constant and the charge on $\mathrm{B}$ remains unchanged
4 the potential difference across $\mathrm{B}$ remains constant while the charge on A remains unchanged
Explanation:
: •When the two capacitor gets charge with same potential the voltage across A \& B and charged is fixed. - When we open the switch capacitor get $\mathrm{A}$ is remains connected with battery but B is isolated so the voltage across A remains constant. And the charge on B is fixed but when we insert the dielectric its capacitance changes so the potential difference also changes.
165713
In the given electrical circuit, if the switch $S$ is closed then the maximum energy stored in the inductors is:
1 $3 \mathrm{~J}$
2 $9 \mathrm{~J}$
3 $12 \mathrm{~J}$
4 $6 \mathrm{~J}$
Explanation:
: Energy in First capacitor, $\mathrm{E}_{1}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ $\mathrm{E}_{1}=\frac{1}{2} \times \frac{4^{2}}{1}=8 \mathrm{~J}$ Energy in second capacitor, $E_{2}=\frac{1}{2} \cdot \frac{Q^{2}}{C}$ $=\frac{1}{2} \cdot \frac{2}{2}^{2}$ $\mathrm{E}_{2} =1 \mathrm{~J}$ Total energy, $\mathrm{E}_{\mathrm{o}}=\mathrm{E}_{1}+\mathrm{E}_{2}$ $=8+1$ $=9 \mathrm{~J}$ Now switch is closed then the common potential of capacitors $\mathrm{V}_{\text {common }}= \frac{\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}_{1}}{\mathrm{C}_{1}}=\frac{4}{1}=4 \mathrm{~V}$ $\mathrm{~V}_{2}=\frac{\mathrm{Q}_{2}}{\mathrm{C}_{2}}=\frac{2}{2}=1 \mathrm{~V}$ $\mathrm{~V}_{\text {common }}=\frac{1 \times 4+2 \times 1}{1+2}$ $=2 \mathrm{~V}$ Hence, the new arrangement of energy, $\mathrm{E}_{1}=\frac{1}{2} \mathrm{C}_{1} \mathrm{~V}_{\text {Common }}^{2}=\frac{1}{2} \times 1 \times 4=2 \mathrm{~J}$ $\mathrm{E}_{2}=\frac{1}{2} \mathrm{C}_{2} \mathrm{~V}_{\text {common }}^{2}=\frac{1}{2} \times 2 \times 4=4 \mathrm{~J}$ Now from conservation of energy stored in the inductor $\mathrm{E}_{1} \square \mathrm{E}_{\mathrm{o}}-\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right)$ $=9-(2+4)$ $=3 \mathrm{~J}$
[AP EAMCET (21.04.2019) Shift-I]
Capacitance
165714
In an oscillating LC circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is
1 $\frac{Q}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{3}}$
3 Q
4 $\frac{\mathrm{Q}}{\sqrt{2}}$
Explanation:
: The electric energy stored when energy is equally distributed. $\mathrm{U}_{\mathrm{E}}^{\prime}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}$ The energy stored in the capacitor when it is maximum charged. $\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ At any time electrostatic energy is equal to magnetic energy. $=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{c}}+\frac{1}{2} \mathrm{LI}^{2}$ $\because \frac{\mathrm{q}^{2}}{2 \mathrm{c}}=\frac{\mathrm{LI}^{2}}{2}$ According to question - $\mathrm{U}_{\mathrm{E}}=2 \mathrm{U}_{\mathrm{E}}^{\prime}$ $\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=2 \times \frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}$ $\frac{\mathrm{Q}^{2}}{2}=\mathrm{q}^{2}$ $\mathrm{q}=\sqrt{\frac{\mathrm{Q}^{2}}{2}}$ $\mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}$
[AP EAMCET (20.04.2019) Shift-1]
Capacitance
165715
The total electrostatic energy stored in both the capacitor is
1 $9 \mu \mathrm{J}$
2 $40.5 \mu \mathrm{J}$
3 $13.5 \mu \mathrm{J}$
4 $18 \mu \mathrm{J}$
Explanation:
: Given, capacitance $\mathrm{C}_{1}=3 \mu \mathrm{F}, \mathrm{C}_{2}=6 \mu \mathrm{F}$ Capacitors are connected in series combination, So, equivalent capacitance $\left(\mathrm{C}_{\mathrm{eq}}\right)=\frac{\mathrm{C}_{1} \times \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ $\mathrm{C}_{\mathrm{eq}}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}$ Since, total electrostatic energy $(U)=\frac{1}{2} C_{\text {eq }} V^{2}$ $\mathrm{U}=\frac{1}{2} \times 2 \times(-3)^{2} \times 10^{-6}$ $\mathrm{U}=9 \mu \mathrm{J}$
[UPSEE - 2017]
Capacitance
165716
Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch $S$. If now the switch $S$ is opened and the space between the plates is filled with dielectric of relative permittivity $\varepsilon_{\mathrm{r}}$, then
1 the potential difference as well as charge on each capacitor goes up by a factor $\varepsilon_{\mathrm{r}}$
2 the potential difference as well as charge on each capacitor goes down by a factor $\varepsilon_{\mathrm{r}}$
3 the potential difference across a remains constant and the charge on $\mathrm{B}$ remains unchanged
4 the potential difference across $\mathrm{B}$ remains constant while the charge on A remains unchanged
Explanation:
: •When the two capacitor gets charge with same potential the voltage across A \& B and charged is fixed. - When we open the switch capacitor get $\mathrm{A}$ is remains connected with battery but B is isolated so the voltage across A remains constant. And the charge on B is fixed but when we insert the dielectric its capacitance changes so the potential difference also changes.
165713
In the given electrical circuit, if the switch $S$ is closed then the maximum energy stored in the inductors is:
1 $3 \mathrm{~J}$
2 $9 \mathrm{~J}$
3 $12 \mathrm{~J}$
4 $6 \mathrm{~J}$
Explanation:
: Energy in First capacitor, $\mathrm{E}_{1}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ $\mathrm{E}_{1}=\frac{1}{2} \times \frac{4^{2}}{1}=8 \mathrm{~J}$ Energy in second capacitor, $E_{2}=\frac{1}{2} \cdot \frac{Q^{2}}{C}$ $=\frac{1}{2} \cdot \frac{2}{2}^{2}$ $\mathrm{E}_{2} =1 \mathrm{~J}$ Total energy, $\mathrm{E}_{\mathrm{o}}=\mathrm{E}_{1}+\mathrm{E}_{2}$ $=8+1$ $=9 \mathrm{~J}$ Now switch is closed then the common potential of capacitors $\mathrm{V}_{\text {common }}= \frac{\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}_{1}}{\mathrm{C}_{1}}=\frac{4}{1}=4 \mathrm{~V}$ $\mathrm{~V}_{2}=\frac{\mathrm{Q}_{2}}{\mathrm{C}_{2}}=\frac{2}{2}=1 \mathrm{~V}$ $\mathrm{~V}_{\text {common }}=\frac{1 \times 4+2 \times 1}{1+2}$ $=2 \mathrm{~V}$ Hence, the new arrangement of energy, $\mathrm{E}_{1}=\frac{1}{2} \mathrm{C}_{1} \mathrm{~V}_{\text {Common }}^{2}=\frac{1}{2} \times 1 \times 4=2 \mathrm{~J}$ $\mathrm{E}_{2}=\frac{1}{2} \mathrm{C}_{2} \mathrm{~V}_{\text {common }}^{2}=\frac{1}{2} \times 2 \times 4=4 \mathrm{~J}$ Now from conservation of energy stored in the inductor $\mathrm{E}_{1} \square \mathrm{E}_{\mathrm{o}}-\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right)$ $=9-(2+4)$ $=3 \mathrm{~J}$
[AP EAMCET (21.04.2019) Shift-I]
Capacitance
165714
In an oscillating LC circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is
1 $\frac{Q}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{3}}$
3 Q
4 $\frac{\mathrm{Q}}{\sqrt{2}}$
Explanation:
: The electric energy stored when energy is equally distributed. $\mathrm{U}_{\mathrm{E}}^{\prime}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}$ The energy stored in the capacitor when it is maximum charged. $\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ At any time electrostatic energy is equal to magnetic energy. $=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{c}}+\frac{1}{2} \mathrm{LI}^{2}$ $\because \frac{\mathrm{q}^{2}}{2 \mathrm{c}}=\frac{\mathrm{LI}^{2}}{2}$ According to question - $\mathrm{U}_{\mathrm{E}}=2 \mathrm{U}_{\mathrm{E}}^{\prime}$ $\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=2 \times \frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}$ $\frac{\mathrm{Q}^{2}}{2}=\mathrm{q}^{2}$ $\mathrm{q}=\sqrt{\frac{\mathrm{Q}^{2}}{2}}$ $\mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}$
[AP EAMCET (20.04.2019) Shift-1]
Capacitance
165715
The total electrostatic energy stored in both the capacitor is
1 $9 \mu \mathrm{J}$
2 $40.5 \mu \mathrm{J}$
3 $13.5 \mu \mathrm{J}$
4 $18 \mu \mathrm{J}$
Explanation:
: Given, capacitance $\mathrm{C}_{1}=3 \mu \mathrm{F}, \mathrm{C}_{2}=6 \mu \mathrm{F}$ Capacitors are connected in series combination, So, equivalent capacitance $\left(\mathrm{C}_{\mathrm{eq}}\right)=\frac{\mathrm{C}_{1} \times \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ $\mathrm{C}_{\mathrm{eq}}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}$ Since, total electrostatic energy $(U)=\frac{1}{2} C_{\text {eq }} V^{2}$ $\mathrm{U}=\frac{1}{2} \times 2 \times(-3)^{2} \times 10^{-6}$ $\mathrm{U}=9 \mu \mathrm{J}$
[UPSEE - 2017]
Capacitance
165716
Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch $S$. If now the switch $S$ is opened and the space between the plates is filled with dielectric of relative permittivity $\varepsilon_{\mathrm{r}}$, then
1 the potential difference as well as charge on each capacitor goes up by a factor $\varepsilon_{\mathrm{r}}$
2 the potential difference as well as charge on each capacitor goes down by a factor $\varepsilon_{\mathrm{r}}$
3 the potential difference across a remains constant and the charge on $\mathrm{B}$ remains unchanged
4 the potential difference across $\mathrm{B}$ remains constant while the charge on A remains unchanged
Explanation:
: •When the two capacitor gets charge with same potential the voltage across A \& B and charged is fixed. - When we open the switch capacitor get $\mathrm{A}$ is remains connected with battery but B is isolated so the voltage across A remains constant. And the charge on B is fixed but when we insert the dielectric its capacitance changes so the potential difference also changes.
165713
In the given electrical circuit, if the switch $S$ is closed then the maximum energy stored in the inductors is:
1 $3 \mathrm{~J}$
2 $9 \mathrm{~J}$
3 $12 \mathrm{~J}$
4 $6 \mathrm{~J}$
Explanation:
: Energy in First capacitor, $\mathrm{E}_{1}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ $\mathrm{E}_{1}=\frac{1}{2} \times \frac{4^{2}}{1}=8 \mathrm{~J}$ Energy in second capacitor, $E_{2}=\frac{1}{2} \cdot \frac{Q^{2}}{C}$ $=\frac{1}{2} \cdot \frac{2}{2}^{2}$ $\mathrm{E}_{2} =1 \mathrm{~J}$ Total energy, $\mathrm{E}_{\mathrm{o}}=\mathrm{E}_{1}+\mathrm{E}_{2}$ $=8+1$ $=9 \mathrm{~J}$ Now switch is closed then the common potential of capacitors $\mathrm{V}_{\text {common }}= \frac{\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}_{1}}{\mathrm{C}_{1}}=\frac{4}{1}=4 \mathrm{~V}$ $\mathrm{~V}_{2}=\frac{\mathrm{Q}_{2}}{\mathrm{C}_{2}}=\frac{2}{2}=1 \mathrm{~V}$ $\mathrm{~V}_{\text {common }}=\frac{1 \times 4+2 \times 1}{1+2}$ $=2 \mathrm{~V}$ Hence, the new arrangement of energy, $\mathrm{E}_{1}=\frac{1}{2} \mathrm{C}_{1} \mathrm{~V}_{\text {Common }}^{2}=\frac{1}{2} \times 1 \times 4=2 \mathrm{~J}$ $\mathrm{E}_{2}=\frac{1}{2} \mathrm{C}_{2} \mathrm{~V}_{\text {common }}^{2}=\frac{1}{2} \times 2 \times 4=4 \mathrm{~J}$ Now from conservation of energy stored in the inductor $\mathrm{E}_{1} \square \mathrm{E}_{\mathrm{o}}-\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right)$ $=9-(2+4)$ $=3 \mathrm{~J}$
[AP EAMCET (21.04.2019) Shift-I]
Capacitance
165714
In an oscillating LC circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is
1 $\frac{Q}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{3}}$
3 Q
4 $\frac{\mathrm{Q}}{\sqrt{2}}$
Explanation:
: The electric energy stored when energy is equally distributed. $\mathrm{U}_{\mathrm{E}}^{\prime}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}$ The energy stored in the capacitor when it is maximum charged. $\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ At any time electrostatic energy is equal to magnetic energy. $=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{c}}+\frac{1}{2} \mathrm{LI}^{2}$ $\because \frac{\mathrm{q}^{2}}{2 \mathrm{c}}=\frac{\mathrm{LI}^{2}}{2}$ According to question - $\mathrm{U}_{\mathrm{E}}=2 \mathrm{U}_{\mathrm{E}}^{\prime}$ $\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=2 \times \frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}}$ $\frac{\mathrm{Q}^{2}}{2}=\mathrm{q}^{2}$ $\mathrm{q}=\sqrt{\frac{\mathrm{Q}^{2}}{2}}$ $\mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}$
[AP EAMCET (20.04.2019) Shift-1]
Capacitance
165715
The total electrostatic energy stored in both the capacitor is
1 $9 \mu \mathrm{J}$
2 $40.5 \mu \mathrm{J}$
3 $13.5 \mu \mathrm{J}$
4 $18 \mu \mathrm{J}$
Explanation:
: Given, capacitance $\mathrm{C}_{1}=3 \mu \mathrm{F}, \mathrm{C}_{2}=6 \mu \mathrm{F}$ Capacitors are connected in series combination, So, equivalent capacitance $\left(\mathrm{C}_{\mathrm{eq}}\right)=\frac{\mathrm{C}_{1} \times \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ $\mathrm{C}_{\mathrm{eq}}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{F}$ Since, total electrostatic energy $(U)=\frac{1}{2} C_{\text {eq }} V^{2}$ $\mathrm{U}=\frac{1}{2} \times 2 \times(-3)^{2} \times 10^{-6}$ $\mathrm{U}=9 \mu \mathrm{J}$
[UPSEE - 2017]
Capacitance
165716
Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch $S$. If now the switch $S$ is opened and the space between the plates is filled with dielectric of relative permittivity $\varepsilon_{\mathrm{r}}$, then
1 the potential difference as well as charge on each capacitor goes up by a factor $\varepsilon_{\mathrm{r}}$
2 the potential difference as well as charge on each capacitor goes down by a factor $\varepsilon_{\mathrm{r}}$
3 the potential difference across a remains constant and the charge on $\mathrm{B}$ remains unchanged
4 the potential difference across $\mathrm{B}$ remains constant while the charge on A remains unchanged
Explanation:
: •When the two capacitor gets charge with same potential the voltage across A \& B and charged is fixed. - When we open the switch capacitor get $\mathrm{A}$ is remains connected with battery but B is isolated so the voltage across A remains constant. And the charge on B is fixed but when we insert the dielectric its capacitance changes so the potential difference also changes.
