If the charge on a parallel-plate capacitor is doubled:

• the capacitance is halved 
• the capacitance is doubled 
• the electric field is halved 
• the electric field is doubled

The capacitance of a parallel-plate capacitor is: 

• proportional to the plate area 
• proportional to the charge stored 
• independent of any material inserted between the plates 
• proportional to the potential difference of the plates 

The capacitance of a parallel-plate capacitor can be increased by: 

• increasing the charge 
• decreasing the charge 
• increasing the plate separation
• decreasing the plate separation 

If both the plate area and the plate separation of a parallel-plate capacitor are doubled, the capacitance is: 

• doubled
• halved
• unchanged 
• one-fourth the original 

A parallel-plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm. To obtain an electric field of 2.0 ´ 10⁶ V/m between the plates, the magnitude of the charge on each plate should be:

• 3.5 ´ 10^-6 C
• 7.1 ´ 10^-6 C 
• 1.4 ´ 10^-5 C
• 1.8 ´ 10^-5 C

What is the voltage across   the plates of the capacitor if  the capacitance is 10 uF and the Charge stored is  30 uC?

• 0.333 V
• 3 V
• 300 V

The figure to the right represents:

• Capacitor
• Diode
• Inductor
• Resistor

With respect to the capacitor, which relationship is true?

• C = Q/V
• V = C×Q
• Q = C/V
• Q = V/C

If the capacitance is 47μF, and the voltage difference across the capacitor terminals is 6V, then the energy stored in that capacitor is:

• 846×10^-6Joule
• 7.23×10^-3 Joule
• 6.76×10^-5Joule
• 5.94×10^-2Joule

Tags

NGSS.HS-PS3-1

If C is the capacitance in Farads, ϵ permittivity of the dielectric, A area of plate overlap in square meters and d distance between plates in meters, then:

• C = ϵd/A 
• C = A/ϵd
• C = ϵA/d
• C = d/ϵA

Tags

NGSS.HS-PS3-5