Question

Two identical parallel conducting plate each with area A having small seperation between them are given charge Q1 and Q2 the magnitude of field between plates is.​

Answer 1

$Answer: \frac{q1 - q2}{2A \epsilon}$

Formula for electric field

$\large \frac{\sigma }{\epsilon}$

where value of Sigma is Positive charge on capacitor divide by Area of plate

But we don't know positive charge on plate

let's find out

Formula for charge=

$\large \frac{Q1-Q2 }{ 2}$

Now coming to final answer

Electric field at P

$\huge \frac{q1 - q2}{2A \epsilon}$

Additional information

• $\large Capacitance= \frac{Q}{V}$
• $\large second \: formula\: for \: capacitance= \frac{\epsilon A}{d}$

Answer 2

Answer:

Electric field within the plates E

Potential difference between the plates

Solution Image (in the attachment)

Explanation:

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