Concept: The charge, Q, stored on a parallel plate capacitor will be equal in magnitude/strength, but opposite in sign (positive charge or negative charge).
To solve for the electric field between the two charged plates of the parallel plate capacitor:
1. Write out the known values.
- Area of one of the plates: 5.68×10-3 m2
- Magnitude of the charge on one of the two plates: 4.38×10-11 C
2. Consider the value we want to solve.
The problem asks for the electric field between the two plates of the parallel plate capacitor. While V=Ed does solve for the electric field, this equation would not consider values such as the charge on one of the plates and so V=Ed wouldn't be as useful a formula. However, another formula you might also recall is:
Ecapacitor = (Q/A) / ε0,
where Q is the charge on one plate, A is the cross sectional area of one plate, and ε0 is the electric constant (8.85×10-12 C2/(Nm2))
3. Apply the equation for the electric field of a parallel plate capacitor:
Recall the knowns from step 1. Solve by plugging in these values into the formula below.
Ecapacitor = (Q/A) / ε0
Note: If you were to simplify the equation above, you may also see it written as Ecapacitor = Q / (A×ε0).
Ecapacitor = (4.38×10-11 C/ 5.68×10-3 m2) / 8.85×10-12 C2/(Nm2)
Ecapacitor = 871.33 N/C
