Let Q represent the charge on a parallel plate capacitor and E the electric field between the plates, then each plate of the capacitor experiences a force of magnitude

Question

Let Q represent the charge on a parallel plate capacitor and E the electric field between the plates, then each plate of the capacitor experiences a force of magnitude (A) zero (B)  QE  (C)  (1/2)QE  (D)  2QE

Tardigrade Admin · Accepted Answer

: Intensity of field E=(q/ε 0A)=(Q/ε 0A) Force between plates =(q2/2ε 0A)=(Q2/2ε 0A) =(q/2)× (q/ε 0A)=(qE/2)=(QE/2)

Q. Let Q represent the charge on a parallel plate capacitor and E the electric field between the plates, then each plate of the capacitor experiences a force of magnitude

zero

$QE$

$\frac{1}{2} QE$

$2 QE$

: Intensity of field $E = \frac{q}{ε _{0} A} = \frac{Q}{ε _{0} A}$ Force between plates $= \frac{q ^{2}}{2 ε _{0} A} = \frac{Q ^{2}}{2 ε _{0} A}$ $= \frac{q}{2} \times \frac{q}{ε _{0} A} = \frac{qE}{2} = \frac{QE}{2}$