Solution Found!
Solved: Two square conducting plates with sides of length
Chapter 24, Problem 77CP(choose chapter or problem)
Problem 77CP
Two square conducting plates with sides of length L are separated by a distance D. A dielectric slab with constant K with dimensions \(L \times L \times D\) is inserted a distance x into the space between the plates, as shown in Fig. P24.77.
(a) Find the capacitance C of this system. (b) Suppose that the capacitor is connected to a battery that maintains a constant potential difference V between the plates. If the dielectric slab is inserted an additional distance dх into the space between the plates, show that the change in stored energy is
\(d U=+\frac{(K-1) \epsilon_{0} V^{2} L}{2 D} d x\)
(c) Suppose that before the slab is moved by \(d x\), the plates are disconnected from the battery, so that the charges on the plates remain constant. Determine the magnitude of the charge on each plate, and then show that when the slab is moved \(d x\) farther into the space between the plates, the stored energy changes by an amount that is the negative of the expression for \(d U\) given in part (b). (d) If F is the force exerted on the slab by the charges on the plates, then \(d U\) should equal the work done against this force to move the slab a distance \(d x\). Thus \(d U=-F d x\). Show that applying this expression to the result of part (b) suggests that the electric force on the slab pushes it out of the capacitor, while the result of part (c) suggests that the force
pulls the slab into the capacitor. (e) Figure 24.16 shows that the force in fact pulls the slab into the capacitor. Explain why the result of part (b) gives an incorrect answer for the direction of this force, and calculate the magnitude of the force. (This method does not require knowledge of the nature of the fringing field.)
Equation Transcription:
Text Transcription:
LXLXD
dx
dU=+(K-1)epsilon_0V^2L over 2D dx
dx
dx
dU
dx
dU=-Fdx
Questions & Answers
QUESTION:
Problem 77CP
Two square conducting plates with sides of length L are separated by a distance D. A dielectric slab with constant K with dimensions \(L \times L \times D\) is inserted a distance x into the space between the plates, as shown in Fig. P24.77.
(a) Find the capacitance C of this system. (b) Suppose that the capacitor is connected to a battery that maintains a constant potential difference V between the plates. If the dielectric slab is inserted an additional distance dх into the space between the plates, show that the change in stored energy is
\(d U=+\frac{(K-1) \epsilon_{0} V^{2} L}{2 D} d x\)
(c) Suppose that before the slab is moved by \(d x\), the plates are disconnected from the battery, so that the charges on the plates remain constant. Determine the magnitude of the charge on each plate, and then show that when the slab is moved \(d x\) farther into the space between the plates, the stored energy changes by an amount that is the negative of the expression for \(d U\) given in part (b). (d) If F is the force exerted on the slab by the charges on the plates, then \(d U\) should equal the work done against this force to move the slab a distance \(d x\). Thus \(d U=-F d x\). Show that applying this expression to the result of part (b) suggests that the electric force on the slab pushes it out of the capacitor, while the result of part (c) suggests that the force
pulls the slab into the capacitor. (e) Figure 24.16 shows that the force in fact pulls the slab into the capacitor. Explain why the result of part (b) gives an incorrect answer for the direction of this force, and calculate the magnitude of the force. (This method does not require knowledge of the nature of the fringing field.)
Equation Transcription:
Text Transcription:
LXLXD
dx
dU=+(K-1)epsilon_0V^2L over 2D dx
dx
dx
dU
dx
dU=-Fdx
ANSWER:
Solution 77CP
Step 1 of 5: