MHT CET · Physics · Capacitance
In parallel plate capacitor, electric field between the plates is ' \(\mathrm{E}\) '. If the charge on the plates is ' \(Q\) ' then the force on each plate is
- A \(\mathrm{QE}\)
- B \(\frac{\mathrm{QE}^2}{2}\)
- C \(\mathrm{QE}^2\)
- D \(\frac{\mathrm{QE}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{QE}}{2}\)
Step-by-step Solution
Detailed explanation
The field produced by charge on each plate is \(\frac{E}{2}\).
Hence force on each plate is given by
\(\mathrm{F}=\mathrm{Q} \times(\) Field produced by the other plate \()\)
\(=\frac{\mathrm{QE}}{2}\)
Hence force on each plate is given by
\(\mathrm{F}=\mathrm{Q} \times(\) Field produced by the other plate \()\)
\(=\frac{\mathrm{QE}}{2}\)
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