2. Consider an ideal dV = 1.5 Volt charged battery connected to (+) and (-) char

Question

2. Consider an ideal dV = 1.5 Volt charged battery connected to (+) and (-) charged AV= 1.5 Volts+ plates, creating a capacitor with electric field vectors of magnitudes E 30 N/C between the plates. The plates are separated by a distance of -5 cm Ignore gravity How much force (in N) is exerted on a q electric field? I- a) +1pCcharge while it is in this In which direction is this force? b) What is the V in Volts) at the (-) plate? What is the V at the (+) plate? (What are the Sl units of "Volts?") c) What is the change in electrical PE n J does it increase or decrease?) when a q +1 charge is put in this electric field, and travels downwards in the electric field from the upper (+) to the lower plate? d) Using your result in (a), how much work (in J) on the q-+1 JC charge willit take to bring it from the e to your answer in (c)? Use the relationship of work e) What is the capacitance of this arrangement if the plates measure 3 cm x 3 cm in area? 1) How much charge is on either plate?

Explanation / Answer

2. given dV = 1.5 V
   electric field inside the capacitor E = 30 N/C
   plate seperation d = 0.05 m

a. force on q = 1 micro C = qE = 1*10^-6*30 = 3*10^-5 N
b. Voltage at + plate = 1.5 V
   Voltage at - plate = 0 V
   SI unit of voltage = Volts, symbol = V
c. when +1 micro C charge moives form +ve to -ve plate, change ion PE = qV = 1*10^-6*1.5 = -1.5*10^-6 J
   the PE of the charge decreases
d. work done to bring charge from -ve to positive plate = -qV = 1.5 *10^-6 J
   this is -ve of the answer from part c
e. capacitance of parallel plate capacitor C = Ae/d
   A = 9 cm^2
   so C = 9*10^-4/4*pi*9*10^9*0.05 = 1.591*10^-13 F
f. charge on either plate = q
   q = CV
   q = 1.591*10^-13*1.5 = 2.387*10^-13 C

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