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Question
A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x=0 and positive plate is at x=3d. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to 3d :
A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x=0 and positive plate is at x=3d. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to 3d :
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Solution
The correct option is C The electric potential increases continuously.
In regions outside the dielectric (regions I and III), the electric field E0 is directed from positive to negative plates.
In region II due to the presence of induced charges, there is an electric field Ek which is opposite to E0, resulting in a net electric field E0−Ek in the direction of E0 (Since E0>Ek).
Hence, the magnitude of the electric field is not the same. However, its direction is from positive to negative plates. Hence, options (a) and (b) are the incorrect options.
Since E0 is directed from the positive to negative plate and we know that potential decreases along the direction of the electric field, electric potential will increase continuously as we go from x=0 to x=3d.
Hence, option (c) is the correct answer.
Alternate Solution:
Even after introduction of dielectric slab, direction of electric field will also be perpendicular to the plates and directed from positive plate to negative plate.
The magnitude of field will be changed.
Before introduced dielectric the field between the plates E=σε0 where, σ is charged density.
After introduced dielectric the field between the plates E=σKε0 , where K is dielectric constant.
As electric field lines flow from higher potential to lower so potential will decrease continuously from x=0 to x=3d.
Hence, option (c) is correct.
Key Concept: The electric field due to induced charges is opposite in direction to the orginal field but lesser in maagnitude.
In regions outside the dielectric (regions I and III), the electric field E0 is directed from positive to negative plates.
In region II due to the presence of induced charges, there is an electric field Ek which is opposite to E0, resulting in a net electric field E0−Ek in the direction of E0 (Since E0>Ek).
Hence, the magnitude of the electric field is not the same. However, its direction is from positive to negative plates. Hence, options (a) and (b) are the incorrect options.
Since E0 is directed from the positive to negative plate and we know that potential decreases along the direction of the electric field, electric potential will increase continuously as we go from x=0 to x=3d.
Hence, option (c) is the correct answer.
Alternate Solution:
Even after introduction of dielectric slab, direction of electric field will also be perpendicular to the plates and directed from positive plate to negative plate.
The magnitude of field will be changed.
Before introduced dielectric the field between the plates E=σε0 where, σ is charged density.
After introduced dielectric the field between the plates E=σKε0 , where K is dielectric constant.
As electric field lines flow from higher potential to lower so potential will decrease continuously from x=0 to x=3d.
Hence, option (c) is correct.
| Key Concept: The electric field due to induced charges is opposite in direction to the orginal field but lesser in maagnitude. |
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