A parallel plate capacitor is constructed of two plates, eachhaving an area of .

Question

A parallel plate capacitor is constructed of two plates, eachhaving an area of .25meters squared. An electric fieldof3.5x10e-6 V/m is establishedwhen the plates are chargedwith42C. a) what voltage is required to charge the capacitor if theplates are spaced .06cm apart? b) What is the dielectric constant(k) of the material betweenthe plates of the capacitor? c) the capacitor is insulated, and the difference between theplates is doubled. what happens to the voltage across thecapacitor? A parallel plate capacitor is constructed of two plates, eachhaving an area of .25meters squared. An electric fieldof3.5x10e-6 V/m is establishedwhen the plates are chargedwith42C. a) what voltage is required to charge the capacitor if theplates are spaced .06cm apart? b) What is the dielectric constant(k) of the material betweenthe plates of the capacitor? c) the capacitor is insulated, and the difference between theplates is doubled. what happens to the voltage across thecapacitor?

Explanation / Answer

We have:
      A = 0.25 m 2       E = 3.5 * 10 -6V / m Charge on the plates q = 42 C                                     = 42 * 10 -6 C a) Separation of the plates d = 0.06 cm                                               = 0.0006 m         Requiredvoltage V = Ed                                       = 2.1 * 10 -9 V b) If differnce in the paltes is doubled capacity is ishalved. Since capacity C = o A / d                         C ( 1/ d ) So, Voltage is doubled

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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