A parallel plate capacitor has an electric fi | vedclass

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(B) The force $F$ on one plate of a parallel plate capacitor due to the electric field $E$ created by the other plate is given by $F = Q 	imes E_{pla

Vedclass · Accepted Answer

$0.05$

Vedclass · Answer

(B) The force $F$ on one plate of a parallel plate capacitor due to the electric field $E$ created by the other plate is given by $F = Q 	imes E_{plate}$.Since the total electric field between the plates is $E = E_{plate} + E_{plate} = 2 E_{plate}$,the field due to one plate is $E_{plate} = E/2$.Therefore,the force on each plate is $F = Q 	imes (E/2) = (Q 	imes E) / 2$.Given $Q = 1 \, \mu C = 10^{-6} \, C$ and $E = 10^5 \, V/m$.Substituting the values: $F = (10^{-6} 	imes 10^5) / 2 = 10^{-1} / 2 = 0.1 / 2 = 0.05 \, N$.

$A$ parallel plate capacitor has an electric field of $10^5 \, V/m$ between the plates. If the charge on the capacitor plate is $1 \, \mu C$,the force on each capacitor plate is ...... $N$.

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