Two plates, each of area 2.50 times 10^-4 m^2, are used to construct a parallel-

Question

Two plates, each of area 2.50 times 10^-4 m^2, are used to construct a parallel-plate capacitor with capacitance 1.00 pF Find the necessary separation distance. If the positive plate is to hold a charge of 5.10 times 10^-12 c, find the charge density. Find the electric field between the plates. What factors determine the electric field strength inside a charged capacitor? what voltage battery should be attached to the plate to obtain the preceding results? What factors determine the potential difference across a capacitor?

Explanation / Answer

Here,

a) let the seperation is d

as capacitance = epsilon * Area/d

1 *10^-12 = 2.50 *10^-4 * 2.5 *10^-4/d

d = 0.00221 m

b)

for the charge density

charge density = charge/area

charge density = 5.1 *10^-12/(2.50 *10^-4)

charge density = 2.04 *10^-8 C/m^2

c)

electric field between the plates = charge density/epsilon

electric field between the plates = 2.04 *10^-8/(8.854 *10^-12)

electric field between the plates = 2304 N/C

d)

voltage across the plates = electric field * d

voltage across the plates = 2304 * 0.002212

voltage across the plates = 5.1 V

the voltage across the plates is 5.1 V

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.