Question

7. + 1/2 points Previous Answers SerPSET9 26.P.041. My Notes Ask Your Teacher The circuit in the figure below consists of two identical, parallel metal plates connected to identical metal springs, a switch, and a 128-V battery. With the switch open, the plates are uncharged, are separated by a distance d = 7.55 mm, and have a capacitance C = 2.00 pF. When the switch is closed, the distance between the plates decreases by a factor of 0.500 (a) How much charge collects on each plate? 512C (b) What is the spring constant for each spring? 2.9 x Your response differs from the correct answer by more than 10%. Double check your calculations. kN/m 6 Show My Work (Optional)

Answer 1

Let,

• $\small Q$ : be the charge on the plate.
• $\small C'$ : be the capacitance after switch is closed
• $\small d'$ : be the distance after switch closed.
• $k$   : be the spring constant.

Given:-

• $\small V = 128 \; V$
• $\small d = 7.55 \; mm = 7.55 \times 10^{-3} \; m$
• $\small C = 2.00 \; \mu F$

After the switch is closed, the new distance between the plates is given by,

$d' = \frac{d}{2} = \frac{7.55}{2}$

$d' =3.775 \; mm$

And the increased in the capacitance is given by,

$C'=2C$

Therefore, the charge on each plate is given by,

$Q = C' V$

$Q = 2C V$

(a) Charge collected on each plate :-

$Q = 2C V$

$Q = 2 \times 6\times 10^{-6} \times 128$

$Q = 512 \times 10^{-6}$

$\mathbf{Q = 512 \; \mu C}$

(b) Spring constant :-

The force on each capacitor plate is given by,

$F_{electrical} = \frac{Q^{2}}{\left (2 \epsilon _{0} \right ) A}$

Where,

• $\small A$ is the area of the plate.
• $\small \epsilon _{0}$ is the permittivity of the free space.

$F_{electrical} = \frac{\left ( 2CV \right )^{2}}{\left (2 \epsilon _{0} \right ) A}$

$F_{electrical} = \frac{4\: C^{2}\: V ^{2}}{2 \left ( \epsilon _{0}\: A \right ) }$

$F_{electrical} = \frac{4\: C^{2}\: V ^{2}}{2 \left (Cd \right ) }$

$F_{electrical} = \frac{2 \: C \: V ^{2}}{d }$

This is the force which pulls the plates together. This must be equal and opposite to the force on each plate due to it attached spring.

Since the plate separation is decreased by the factor of 0.500, each spring will have decreased length by d/4 ,therefore, the restoring force is given by,

$F_{spring} = - k \left (\frac{d}{4} \right )$

Therefore,

$F_{spring} + F_{electrical} = 0$

$- k \left (\frac{d}{4} \right ) + \frac{2 \: C \: V ^{2}}{d } = 0$

$k = \frac{8 \: C \: V ^{2}}{d^{2} }$

$k = \frac{8 \left ( 2\times 10^{-6} \right ) \: \left ( 128 \right ) ^{2}}{ \left ( 7.55\times 10^{-3} \right )^{2} }$

$k = \frac{0.2621 }{ \left (57 \times 10^{-6} \right ) }$

$\mathbf{k = 4.6 \times 10^{3} \; N/m}$

or

$\mathbf{k = 4.6 \; kN/m}$

This is the value of the spring constant for each spring.