Consider two horizontal parallel plates. The upper plate has a voltage differenc

Question

Consider two horizontal parallel plates. The upper plate has a voltage difference of 30 V with the lower plate, and they are separated by 3 cm. You move the two plates carefully to a separation of 4 cm. What is the strength of the electric field between the plates?
A. 3000 volts/meter
B. 1000 volts/meter
C. 750 volts/meter
D. 7.5 volts/meter
E. 75 volts/meter

If a charge of -3 x 10-6 C were allowed to fall through a potential difference of +500 V, the change in potential energy for the charge would be
A. -1.5

Explanation / Answer

electric field remain unchanged in capacitor is battery is not connected

so the electric field is E

E = V / d

E = 30 / 3 * 10^-2

E = 1000 V/m

potential energy = q * V

potential energy = -3 * 10^-6 * 500

potential energy = -1.5 * 10^-3 J

negative amount of work is to be done to bring a negative charge from an infinitely great distance away into the presence of a positive charge

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects