SATHEE: Electrostatics Question 277

Electrostatics Question 277

Question: In an isolated parallel plate capacitor of capacitance C, the four surface have charges $Q_{1}$ , $Q_{2}$ , $Q_{3}$ and $Q_{4}$ as shown. The potential difference between the plates is [IIT-JEE 1999]

Options:

A) $400 V$

B) $800 V$

C) $1200 V$

D) $1600 V$

Show Answer

Answer:

Correct Answer: B

Solution:

Charge on capacitor A is given by

$Q_{1} = 15 \times 10^{- 6} \times 100 = 15 \times 10^{- 4} C$

Charge on capacitor B is given by $Q_{2} = 1 \times 10^{- 6} \times 100 = 10^{- 4} C$

Capacity of capacitor A after removing dielectric $= \frac{15 \times 10^{- 6}}{15} = 1 μ F$

Now when both capacitors are connected in parallel their equivalent capacitance will be Ceq $= 1 + 1 = 2 μ F$

So common potential $= \frac{(15 \times 10^{- 4}) + (1 \times 10^{- 4})}{2 \times 10^{- 6}} = 800 V .$

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Electrostatics Question 277

Question: In an isolated parallel plate capacitor of capacitance C, the four surface have charges Q1 , Q2 , Q3 and Q4 as shown. The potential difference between the plates is [IIT-JEE 1999]

Options:

Answer:

Solution:

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Question: In an isolated parallel plate capacitor of capacitance C, the four surface have charges $Q_{1}$ , $Q_{2}$ , $Q_{3}$ and $Q_{4}$ as shown. The potential difference between the plates is [IIT-JEE 1999]