A 11.0 \\mu F parallel-plate capacitor with circular plates is connected to a 10

Question

A 11.0 mu F parallel-plate capacitor with circular plates is connected to a 10.0 V battery.
Part A
What is the charge on each plate?
q1 = ____________µC
Part B
How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery?
q2 = ______________µC
Part C
How much charge would be on the plates if the capacitor were connected to the 10.0 V battery after the radius of each plate was doubled without changing their separation?
q3 = _____________µC

Explanation / Answer

Capacitance C = 11.0 F = 11 * 10 ^-6 F voltage V = 10.0 volt (1). charge on each plate Q = CV = 110.0*10^-6 coulomb                                            = 110.0 coulomb (2). The capacitor remian connected to battery.So, voltage remains constant. we know C = A / d        ----( 1)                 C 1/ d        C' / C = d / d'                  = d /(2d)                  = 1/ 2             C ' = C / 2 charge on the plates if their separation were doubled while the capacitor remained connected to the battery is Q ' = C ' V = 55.0 coulomb (3). from eq( 1) , C A    C ' / C = A ' / A
              = ( r' / r) 2     Since area A = r^ 2               = ( 2r / r) ^ 2               = 2^ 2               = 4            C ' = 4C charge on the plates if the capacitor were connected to the battery after the radius of each plate was doubled without changing their separation is Q " = C ' * V = 4CV = 440.0 coulomb
Capacitance C = 11.0 F = 11 * 10 ^-6 F voltage V = 10.0 volt (1). charge on each plate Q = CV = 110.0*10^-6 coulomb                                            = 110.0 coulomb (2). The capacitor remian connected to battery.So, voltage remains constant. we know C = A / d        ----( 1)                 C 1/ d        C' / C = d / d'                  = d /(2d)                  = 1/ 2             C ' = C / 2 charge on the plates if their separation were doubled while the capacitor remained connected to the battery is Q ' = C ' V = 55.0 coulomb (3). from eq( 1) , C A    C ' / C = A ' / A
              = ( r' / r) 2     Since area A = r^ 2               = ( 2r / r) ^ 2               = 2^ 2               = 4            C ' = 4C charge on the plates if the capacitor were connected to the battery after the radius of each plate was doubled without changing their separation is Q " = C ' * V = 4CV = 440.0 coulomb

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.