A parallel-plate capacitor, with air between the plates, is connected across a v

Question

A parallel-plate capacitor, with air between the plates, is connected across a voltage source. This source establishes a potential difference between the plates by placing charge of magnitude 3.64 ×10 -6 C on each plate. The space between the plates is then filled with a dielectric material, with a dielectric constant of 7.74. What must the magnitude of the charge on each capacitor plate now be, to produce the same potential difference between the plates as before?

2.82×105 C

3.22×105 C

1.03×104 C

3.64×106 C

4.7×107 C

Explanation / Answer

let the voltage source be of V volts

chagre = V*C=3.64*10^(-6) F

if dielectri of 7.74 is kept Cnew=C*7.74=7.74C

for same potential V .let charge be Q

Q=Cnew*V

Q=7.74*3.64*10^(-6) =2.81736*10^(-5) F

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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