An electrical capacitor with two parallel plates, has a capacitance of 4 micro farads, if it was half filled with a dielectric with a Permittivity of 2 ε, its new capacitance in micro farad is:
a) 6
b) 3
c) 4
d) 2
An electrical capacitor with two parallel plates, has a capacitance of 4 micro farads, if it...
A parallel plate capacitor has plates of area A = 5.50 ✕ 10−2 m2 separated by distance d = 1.32 ✕ 10−4 m. (The permittivity of free space is ε0 = 8.85 ✕ 10−12 C2/(N · m2).) (a) Calculate the capacitance (in F) if the space between the plates is filled with air. . What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant κ = 3.10 as in...
A capacitor has a capacitance of 49 micro-farads without a dielectric while connected to a 20 volt battery. Then the distance between the plates is decreased by a factor of 5 and a dielectric with a dielectric constant of 2.29 is inserted between the plates of the capacitor while still connected to the same battery. What is voltage, in volts, across the capacitor after the dielectric and the distance between the plates have been changed?
A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them. When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates. While it is connected to the battery the space between the plates is filled with a material of dielectric constant k=3. After the dielectric is added, the magnitude of the charge on the plates and the new capacitance are
Problem 5 (16 +8 Points) A parallel plate capacitor biased under a constant voltage Vo is shown in the figure below. Two parallel plates are separated by a distance of d and filled with a medium whose permittivity is ε,-ε-The parallel plates have a height of h and a width of t not shown in the figure). A dielectric slab with a thickness of d/3 is inserted in between the two parallel plates. The dielectric slab has a permittivity of...
A parallel plate capacitor of capacitance Co has plates of area A with separation d between them. When it is connected to a battery of voltage Vo, it has a charge of magnitude Qo on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitudes of the charge on the plates and the potential difference between them are 15.
An air-filled parallel-plate capacitor has a capacitance of 1.2 pF. The separation of the plates is doubled and wax is inserted between them. The new capacitance is 1.8 pF. Find the dielectric constant of the wax.
An air-filled parallel-plate capacitor has a capacitance of 1.4 pF. The separation of the plates is doubled and wax is inserted between them. The new capacitance is 2.1 pF. Find the dielectric constant of the wax.
A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them. When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates. The plates are pulled apart to a separation 2d while the capacitor remains connected to the battery and the space between the plates is filled halfway with a material having the dielectric constant K. What are the capacitance and the magnitude of the charge...
A 2.5 nF parallel-plate capacitor has an air gap between its plates. Its capacitance increases by 2.5 nF when the gap is filled by a dielectric What is the dielectric constant of that dielectric?
A parallel plate capacitor has plates of area A-5.50 × 10-2 m2 separated by distance d 1.77 x 10 4 m. hepe mittivity of e e space is e 8.85x 1 Hr N-m. HINT (a) Calculate the capacitance (in F) if the space between the plates is filled with air. What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant K-3.30 as in figure (a), and figure (b)? (Hint:...