Questions
A parallel plate capacitor of plate area A and plate separation d is charged to potential V and then the battery is disconnected . A slab of dielectric constant K is then inserted between the plates of the capacitors so as to fill the space between the plates . If Q, E and W denote respectively the magnitude of charge on Each plate, the electric field between the plates (after the slab inserted) and work done on the system in question in the process of inserting the slab , then state incorrect from the following:
detailed solution
Correct option is B
After inserting the dielectric slabNew capacitance C'=K.C=Kε0AdNew potential difference V'=VKNew charge Q'=C'V'=ε0AVd Charge remains same.New electric filed E'=V'd=VKdWork done (W)= Final energy – initial energyW=12C'V'2−12CV2=12KCVK2−12CV2 =12CV21K−1=−12CV21−1K =−ε0AV22d1−1K ⇒W=ε0AV22d1−1KTalk to our academic expert!
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The plates of parallel plate capacitor are charged upto 100 V. A 2 mm thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the plate is
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