A C1= 2.50uF capacitor is charged to 862 V and a C2= 6.80uF capacitor is charged to 660 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each? [Hint: charge is conserved.] What will be the charge on each?

Question

A  C1= 2.50uF capacitor is charged to 862 V and a  C2= 6.80uF capacitor is charged to 660 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other.
What will be the potential difference across each? [Hint: charge is conserved.]
What will be the charge on each?

Damon · Accepted Answer

C = Q/V  (That is the meaning of C)

how much charge on 2.5*10^-6 at 862 
Q1 = C V =  2.5*10^-6 * 862 = 2155*10^-6 coulombs

how much charge on 6.8*10^-6 at 660 ?
Q2 = 6.8*10^-6 * 660 = 4488*10^-6

Total charge = Q1+Q2 = 6643*10^-6 
Total Capacitance = (2.5+6.8)10^-6 =  9.30*10^-6 Farads

V = same on each now = 6643/9.30 = 714.3 volts

now just do
Q1 = C1 (714)
Q2 = C2 (714
for the charge on each

A C1= 2.50uF capacitor is charged to 862 V and a C2= 6.80uF capacitor is charged to 660 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other.

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