A charged capacitor stores energy u. Without connecting this capacitor to anythi

Question

A charged capacitor stores energy u. Without connecting this capacitor to anything dielectric having dielectric constant K is now inserted between the plates of the capacitor, completely filling the space between them. How much energy does the Equal but opposite charges Q are placed on the square plates capacitor. the plates are then pulled sprat to twice there original compared to the direction of the plates. Which of the following capacitor are true? the energy stored in the capacitor has doubled. the potential difference across the plates has increased. the capacitance has doubt the energy density in the capacitor has increased/Twp capacitors.C_1 and C_2 are connected in series across a source potential source still connected, a dielectric in now What happens to the charge on capacitor C_2 the charge on C_2 decreases. the charge on C_2 increases. the charge on C_2 decreases.

Explanation / Answer

4) The energy stored in the capacitor is given by U = Q2/(2C) where Q is the stored charge of the capacitor and C is its capacitance. Putting a dielectric between the capacitor plates changes C to KC and thus U to U/K (option B)

5) A) (True) the energy stored in the capacitor has doubled as U = Q^2/2C [C = e0A/d]

B) (True) the potential difference across the plates has doubled as V = Q/C

C) (False) Electric field between the plates has been increased as E = V/d

D) (False) Because capacitor will be halved

E) (False) As u = U/V ===> U and V both get double so u remain unchanged.

6)   C1's value now increases. That means the total capacitance of the two in series also increases. Since the voltage is fixed, the charge increases, Q = CV.
And that charge is on both caps equally, so the answer is (B)

7) U = kq1q2/r = (9*10^9*3*10^-6*9*10^-6)/(6*10^-3) = 40.5 J (option D)

8)  As C = Q/V (Charge/Voltage) and
C=eA/d (dielectric constant*Area/seperation of the plates)
When you put that dielectric in (change the second equation), you increase the capacitance of the capacitor. The voltage (of the first equation) is still the same, so that means more charge will start to flow into the capacitor.

Option D is correct.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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