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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 28 of 37
Marks: +1, -0
Show that the force on each plate of a parallel plate capacitor has magnitude equal to 12\frac{1}{2} QE , where Q is the charge on the capacitor, and E→\overset{\rightarrow}{E} is the magnitude of electric field between the plates. Explain the origin of the factor 12\frac{1}{2}
Solution:  
Magnitude of electric field between the plates of charged capacitor is E = σε0\frac{\sigma}{\varepsilon_0}
However, magnitude of electric field of one plate on the other plate of charged capacitor is E1E_1 = σ2ε0\frac{\sigma}{2\varepsilon_0} . So, force on the one plate of charged capacitor due to the other is
F = QE1QE_1 = Q . σ2ε0\frac{\sigma}{2\varepsilon_0} = 12\frac{1}{2} × Q × σε0\frac{\sigma}{\varepsilon_0} or F = 12\frac{1}{2} QE
The factor 12\frac{1}{2} is because the electric field of one plate on the other plate of charged capacitor is 12\frac{1}{2} of the resultant electric field E between the plates of charged capacitor.
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