The electric field between the plates of a parallel plate capacitor when connected to a certain battery is E if the space between the plates of the capacitor is filled by introducing a material of dielectric constant k without disturbing the battery connections the electric field between the plates shall be?

Question

Answer

The main thing is that the 'battery is still connected'. So, the potential difference across the plates doesn't change even when the dielectric slab is introduced. On introducing the slab with dielectric k the capacitance of the capacitor increases (by k times) and to keep the potential difference constant the charge increases (by k times). Since electric field, E = V/d ther will be no change in the field too.

If the question is such that the battery is removed before inserting the slab then charge would remain constant. As the capacitance increases by k times potential decreases by k times and so is the electric field.

Answer

Introduction:
When a parallel plate capacitor is connected to a certain battery, the electric field between the plates is E. Now, if a material of dielectric constant k is introduced between the plates of the capacitor without disturbing the battery connections, the electric field between the plates will change.

Effect of Dielectric on Electric Field:
When a dielectric material is introduced between the plates of a capacitor, the electric field between the plates decreases. The new electric field between the plates can be calculated using the formula:

E' = E/k

Where E' is the new electric field, E is the initial electric field, and k is the dielectric constant of the material introduced.

Explanation:
The reason for the decrease in the electric field is due to the polarization of the dielectric material. When a dielectric material is introduced between the plates of a capacitor, the electric field causes the electrons in the material to shift slightly, creating a dipole moment. These dipoles produce an electric field in the opposite direction to the initial electric field, thus reducing the net electric field between the plates.

Example:
Suppose a parallel plate capacitor is connected to a battery, and the electric field between the plates is 100 N/C. If a dielectric material with a dielectric constant of 2 is introduced between the plates without disturbing the battery connections, the new electric field between the plates would be:

E' = 100/2 = 50 N/C

Thus, the introduction of the dielectric material reduces the electric field between the plates by a factor of 2.

Conclusion:
In conclusion, the introduction of a dielectric material between the plates of a capacitor decreases the electric field between the plates. The new electric field can be calculated using the formula E' = E/k, where E is the initial electric field, k is the dielectric constant of the material introduced, and E' is the new electric field.

Similar Class 12 Doubts

26. (

Two identical parallel plate capacitors are connected in series and then joined with a battery of 100V. A sheet of dielectric constant 4 is inserted between the plates of second capacitor.?

A Capacitor stores 50microC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 500microC flows through the battery . What is the dielectric constant used?

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