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The plates of a parallel plate capacitor are given charges +4Q and _2Q. The capacitor is then connected across an uncharged capacitor of same capacitance as first one (= C). Find the final potential difference between the plates of the first capacitor.
Correct answer is '3Q/2C'. Can you explain this answer?
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Verified Answer
The plates of a parallel plate capacitor are given charges +4Q and _2Q...
Based on the symmetry of charges at equilibrium condition, +4Q will divide into −2Q each. Similarly−2Q will divide into −Q each.
Effectively capacitance=C+C=2C
So, the final potential difference = 2Q−(Q)/ 2C =3Q/2C
Effectively capacitance=C+C=2C
So, the final potential difference = 2Q−(Q)/ 2C =3Q/2C
Most Upvoted Answer
The plates of a parallel plate capacitor are given charges +4Q and _2Q...
Explanation:
When the parallel plate capacitor with charges 4Q and -2Q is connected across an uncharged capacitor of the same capacitance, the charges redistribute themselves to attain a common potential across the two capacitors.
Step 1: Calculation of equivalent capacitance
The equivalent capacitance of the two capacitors in parallel is given by:
1/Ceq = 1/C + 1/C
Ceq = C/2
Step 2: Calculation of charge on the equivalent capacitor
The charge on the equivalent capacitor is the sum of the charges on the two capacitors in parallel.
Qeq = Q1 + Q2
Qeq = 4Q - 2Q
Qeq = 2Q
Step 3: Calculation of potential difference
The potential difference across the plates of the equivalent capacitor is given by:
Veq = Qeq/Ceq
Veq = 2Q/(C/2)
Veq = 4Q/C
The potential difference across the plates of the first capacitor is the same as the potential difference across the plates of the equivalent capacitor, since they are connected in parallel.
V1 = Veq
V1 = 4Q/C
Step 4: Simplification of the answer
The final potential difference between the plates of the first capacitor is:
V1 = 4Q/C
V1 = (4/2)(Q/C)
V1 = 2Q/C
V1 = (3/2)(2Q/C)
V1 = 3Q/2C
Therefore, the final potential difference between the plates of the first capacitor is 3Q/2C.
When the parallel plate capacitor with charges 4Q and -2Q is connected across an uncharged capacitor of the same capacitance, the charges redistribute themselves to attain a common potential across the two capacitors.
Step 1: Calculation of equivalent capacitance
The equivalent capacitance of the two capacitors in parallel is given by:
1/Ceq = 1/C + 1/C
Ceq = C/2
Step 2: Calculation of charge on the equivalent capacitor
The charge on the equivalent capacitor is the sum of the charges on the two capacitors in parallel.
Qeq = Q1 + Q2
Qeq = 4Q - 2Q
Qeq = 2Q
Step 3: Calculation of potential difference
The potential difference across the plates of the equivalent capacitor is given by:
Veq = Qeq/Ceq
Veq = 2Q/(C/2)
Veq = 4Q/C
The potential difference across the plates of the first capacitor is the same as the potential difference across the plates of the equivalent capacitor, since they are connected in parallel.
V1 = Veq
V1 = 4Q/C
Step 4: Simplification of the answer
The final potential difference between the plates of the first capacitor is:
V1 = 4Q/C
V1 = (4/2)(Q/C)
V1 = 2Q/C
V1 = (3/2)(2Q/C)
V1 = 3Q/2C
Therefore, the final potential difference between the plates of the first capacitor is 3Q/2C.
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Community Answer
The plates of a parallel plate capacitor are given charges +4Q and _2Q...
Based on the symmetry of charges at equilibrium condition, +4Q will divide into −2Q each. Similarly−2Q will divide into −Q each.
Effictively capacitance=C+C=2C
So, the final potential difference =2C2Q−(Q)=2C3Q
Effictively capacitance=C+C=2C
So, the final potential difference =2C2Q−(Q)=2C3Q
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The plates of a parallel plate capacitor are given charges +4Q and _2Q. The capacitor is then connected across an uncharged capacitor of same capacitance as first one (= C). Find the final potential difference between the plates of the first capacitor.Correct answer is '3Q/2C'. Can you explain this answer?
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