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PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Two identical metal plates are given positive charges $Q_{1}$ and $Q_{2} (< Q_{1})$ respectively. If they are now brought close together to form a parallel plates capacitor with capacitance $C$ , the potential difference between them is

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Answer:

Solution:

For charge $Q_{1}$ , electric field is

$E_{1} = \frac{Q_{1}}{2 ε_{0} A}$

where $A$ is area.

For charge $Q_{2}$ electric field is

$E_{2} = \frac{Q_{2}}{2 ε_{0} A}$

Resultant electric field

$E = E_{1} - E_{2} = \frac{(Q_{1} - Q_{2})}{2 ε_{0} A}$

Potential difference between plates when they are brought close together to form a parallel plate capacitor is

$V = E d = \frac{(Q_{1} - Q_{2})}{2 ε_{0} A} d$

Since, $C = \frac{ε_{0} A}{d}$ for parallel plate capacitor.

$V = \frac{(Q_{1} - Q_{2})}{2 C}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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