BATTERY after charging, the charge stored in eke capacitor remains constant. When a capacitor is CHARGED by connecting a battery across its plates, the initial energy stored, \[U=\frac{{{q}^{2}}}{2C}\] When the battery is DISCONNECTED, then the charge remains constant i.e., \[q=\]constant. Now another identical capacitor is connected across it i.e., the capacitors are connected in PARALLEL, so the equivalent capacitance \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=C+C=2C\] Thus, final energy stored by the system of capacitors, \[U'=\frac{{{q}^{2}}}{2{{C}_{eq}}}\] \[=\frac{{{q}^{2}}}{2(2C)}=\frac{1}{2}U\] \[\THEREFORE \] \[U'=\frac{U}{2}\]

"> BATTERY after charging, the charge stored in eke capacitor remains constant. When a capacitor is CHARGED by connecting a battery across its plates, the initial energy stored, \[U=\frac{{{q}^{2}}}{2C}\] When the battery is DISCONNECTED, then the charge remains constant i.e., \[q=\]constant. Now another identical capacitor is connected across it i.e., the capacitors are connected in PARALLEL, so the equivalent capacitance \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=C+C=2C\] Thus, final energy stored by the system of capacitors, \[U'=\frac{{{q}^{2}}}{2{{C}_{eq}}}\] \[=\frac{{{q}^{2}}}{2(2C)}=\frac{1}{2}U\] \[\THEREFORE \] \[U'=\frac{U}{2}\]

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A capacitor is charged by connecting a battery across its plates. It stores energy U. Now the battery is disconnected and another identical capacitor is connected across it, then the energy stored by both capacitors of the system will be: [AIPMT 2000]

NEET Physics in NEET . 3 months ago

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[b] Key Idea: On removing the BATTERY after charging, the charge stored in eke capacitor remains constant. When a capacitor is CHARGED by connecting a battery across its plates, the initial energy stored, \[U=\frac{{{q}^{2}}}{2C}\] When the battery is DISCONNECTED, then the charge remains constant i.e., \[q=\]constant. Now another identical capacitor is connected across it i.e., the capacitors are connected in PARALLEL, so the equivalent capacitance \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=C+C=2C\] Thus, final energy stored by the system of capacitors, \[U'=\frac{{{q}^{2}}}{2{{C}_{eq}}}\] \[=\frac{{{q}^{2}}}{2(2C)}=\frac{1}{2}U\] \[\THEREFORE \] \[U'=\frac{U}{2}\]

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