Initial charges with proper sign on the plates of two identical capacitors - each of 1 - F- are as shown- When both S1 and S2 are closed- the potential difference between A and B in volts will finally become -

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Byju's Answer

Standard XII

Physics

Capacitance of Parallel Plate Capacitor

Initial charg...

Question

Initial charges (with proper sign) on the plates of two identical capacitors , each of $1 μ F$ , are as shown. When both $S_{1}$ and $S_{2}$ are closed, the potential difference between A and B in volts will finally become :

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Solution

The potential difference across the two capacitors is the same. Let the charge across capacitor 1 be q . The charge across capacitor 2 is 80-q . Since the capacitances are same,
$q = 80 - q \Rightarrow q = 40 μ C$
Potential difference $= q / C = 40 V$

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Initial charges (with proper sign) on the plates of two identical capacitors , each of 1μF, are as shown. When both S1 and S2 are closed, the potential difference between A and B in volts will finally become :

The potential difference across the two capacitors is the same. Let the charge across capacitor 1 be q . The charge across capacitor 2 is 80-q . Since the capacitances are same,q=80−q⇒q=40μCPotential difference =q/C=40V

Initial charges (with proper sign) on the plates of two identical capacitors , each of $1 μ F$ , are as shown. When both $S_{1}$ and $S_{2}$ are closed, the potential difference between A and B in volts will finally become :

The potential difference across the two capacitors is the same. Let the charge across capacitor 1 be q . The charge across capacitor 2 is 80-q . Since the capacitances are same,
$q = 80 - q \Rightarrow q = 40 μ C$
Potential difference $= q / C = 40 V$