A distance of 50 cm separates the two metal plates of a parallel plate capacitor

Question

A distance of 50 cm separates the two metal plates of a parallel plate capacitor. The plates are charged, creating a constant electric field between them. A 0.3 Coulomb charged particle is placed in the space between the plates. If the force on this charged particle is 20 N, what is the potential difference between the plates? 3.3 V 6 V 16.7 V 33 V There is no potential difference since E is constant For the figure shown below, q_1 = 4 times 10^-6 C is 2 m from the origin on the x-axis and q_2 = .l times 10^-6 Cis 1 m from the origin on the y-axis, as shown. What is the magnitude and direction of the net dearie field at the origin? 0 N/C 12,700 N/C 45 degrees above x-axis 12,700 N/C 45 degrees below x-axis 18000 N/C 45 degrees above x-axis 18000 N/C 26.5 degrees above x-axis

Explanation / Answer

Field inside capacitor will be

E = Q / e0 A

where Q is teh charge on plates and A is area of plate.

and F = q E

20 = 0.3 (Q / e0 A )

Q / e0A = 66.67

and capacitance , C = e0 A / d

and Q = C V

Q = (e0 A / d) ( V)

V = (Q / e0 A) d = 66.67 x 0.50 = 33.3 V

ANs(D)

9. field due to a point charge at distance r , E= kq/r^2

due to q1, E1 = (9 x 10^9 x 4 x 10^-6) / (2^2) (i)

E1 = 9000i N/C

due to q2: E2 = (9 x 10^9 x 1 x 10^-6)/1^2 j = 9000 j N/C

Enet = E1 + E2 = 9000i + 9000j N/C

magniude = sqrt(9000^2 + 9000^2) = 12727 N/C

directiona = tan^-1(9000/9000) = 45 deg

Ans (B)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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