Potential difference(capacitor)

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Homework Statement



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Homework Equations





The Attempt at a Solution



Worked calculation of capacitance and plate charge: C = (epsilon_0 A)/d, Q = CV, with epsilon_0 = 8.85 x 10^-12, A = 0.50, d = 25 x 10^-3, V = 75 - (-75) C, giving Q = 2.66 x 10^-8 C.


1) "One plate is maintained at +75 V while the other at -75 V by a d.c. supply" means? Does it means that this capacitor is connected to a 150 V d.c. supply or just a 75 V?
2) Why is 75-(-75)? Why don't we just use 75V?
3) Charge stored on each plate is it the same as charge stored in the whole capacitor?
:confused:
 
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  • Printed exam question: two thin conducting plates, each 0.50 m² in area, 25 mm apart, one at +75 V and the other at -75 V from a d.c. supply; part (i) asks for the charge on each plate, 4 marks.
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  • Worked solution for capacitance and charge: C = (epsilon_0 A)/d, Q = CV, then Q = (8.85 x 10^-12)(0.50)/(25 x 10^-3) [75 - (-75)] C = 2.66 x 10^-8 C.
    Capture1.JPG
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stpmmaths said:
1) "One plate is maintained at +75 V while the other at -75 V by a d.c. supply" means? Does it means that this capacitor is connected to a 150 V d.c. supply or just a 75 V?
All that matters with capacitors is the potential difference across the plates because that is the voltage V as in Q = C V. So you can view this as one plate being at potential 0 V and the other at 150 V or -75 V and 75 V, or 250 V and 100 V.
2) Why is 75-(-75)? Why don't we just use 75V?
See above. What is the difference in electrostatic potential between the plates?
3) Charge stored on each plate is it the same as charge stored in the whole capacitor?
:confused:

One plate has positive charge (+Q) and the other plate has negative charge of the same magnitude (-Q). By convention, when we say "the charge on the capacitor", we mean the charge on the positive plate, +Q.