A dielectric slab (εr = 4) is inserted fully into a charged parallel plate capacitor with constant charge. The electric field between the plates:
Correct answer: B. Decreases by a factor of 4
- A. Increases by a factor of 4
- B. Decreases by a factor of 4
- C. Remains unchanged
- D. Becomes zero
Explanation
With constant charge, E = Q/(ε₀εrA). Inserting a dielectric with εr = 4 reduces E by a factor of 4, as the electric field is inversely proportional to the dielectric constant.
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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