An air-filled parallel plate capacitor is charged until its surface charge density reaches σ0, and then the voltage source is removed. If an insulator with dielectric constant 5.0 is inserted between the plates, what happens to the electric field strength, assuming the surface charge density does not change?

Question

Quizplus · Accepted Answer

The electric field strength decreases by a factor of 5 because the introduction of the insulator increases the capacitance of the capacitor by a factor of 5 (since C = εA/d, where ε is the dielectric constant, A is the area of the plates, and d is the distance between the plates). Since Q = σ0A, where Q is the charge on each plate, the voltage, V, is given by V = Q/C. Therefore, when C increases by a factor of 5, V decreases by a factor of 5. Since the electric field strength is proportional to the voltage (E = V/d), it also decreases by a factor of 5.

An Air-Filled Parallel Plate Capacitor Is Charged Until Its Surface