Chapter 17, Problem 49P

(a)

To determine

The capacitance will be estimated.

Answer to Problem 49P

The answer is $7.265\times {10}^{-12}F$ .

Explanation of Solution

Given data:

Distance between plates, $d=5.0\times {10}^{-2}m$

Radius, $4.5inch\times 0.254m/inch$

Formula used:

It is known that: $C=\frac{{\epsilon }_{0}A}{d}$

Where,

  $\text{C}$ is the capacitance.

  $A$ is the area of the plate.

  $d$ is the distance between plates.

Calculation:

Applying the formula to calculate the capacitance, $C=\frac{{\epsilon }_{0}A}{d}$

  $\therefore C=\frac{\left(8.85\times {10}^{-12}{C}^2/N\cdot m^2\right)\pi {\left(4.5inch\times 0.254m/inch\right)}^2}{5.0\times {10}^{-2}m}$

  $\therefore C=7.265\times {10}^{-12}F$

Conclusion: The required capacitance is $7.265\times {10}^{-12}F$ .

(b)

To determine

The charge on each plate will be estimated.

Answer to Problem 49P

The answer is $6.54\times {10}^{-11}C$ .

Explanation of Solution

Given data:

Capacitance, $C=7.265\times {10}^{-12}F$

Voltage, $V=9V$

Formula used:

It is known that: $Q=CV$

Where,

  $\text{C}$ is the capacitance.

  $Q$ is charge.

  $V$ is the potential.

Calculation:

Applying the formula to calculate the charge, $Q=CV$

  $\therefore Q=\left(7.2659\times {10}^{-12}F\right)\left(9V\right)$

  $\therefore Q=6.54\times {10}^{-11}C$

Conclusion: The required charge is $6.54\times {10}^{-11}C$ .

(c)

To determine

The electric field halfway between the plates will be estimated.

Answer to Problem 49P

The answer is $180V/m$ .

Explanation of Solution

Given data:

Distance between plates, $d=5.0\times {10}^{-2}m$

Voltage, $V=9V$

Formula used:

It is known that: $E=\frac{V}{d}$

Where,

  $\text{E}$ is the electric charge.

  $d$ is the distance between the plates.

  $V$ is the potential.

Calculation:

Applying the formula to calculate the electric field, $E=\frac{V}{d}$

  $\therefore E=\frac{\left(9V\right)}{5.0\times {10}^{-2}m}$

  $\therefore E=180V/m$

Conclusion: The required electric field is $180V/m$ .

(d)

To determine

The work done by the battery is calculated.

Answer to Problem 49P

The answer is $2.94\times {10}^{-10}J$ .

Explanation of Solution

Given data:

Charge, $Q=6.54\times {10}^{-11}C$

Voltage, $V=9V$

Formula used:

It is known that: $PE=\frac{QV}{2}$

Where,

  $PE$ is the stored energy.

  $Q$ is charge.

  $V$ is the potential.

Calculation:

Applying the formula to calculate the charge, $PE=\frac{QV}{2}$

  $\therefore PE=\frac{\left(6.54\times {10}^{-11}C\right)\left(9V\right)}{2}$

  $\therefore PE=2.94\times {10}^{-10}J$

Conclusion: The work done by the battery is $2.94\times {10}^{-10}J$ .

(e)

To determine

The value that gets changed if a dielectric is inserted

Answer to Problem 49P

The following things will changes when a dielectric is inserted.

• Capacitance
• Charge
• Energy

Explanation of Solution

When a dielectric is inserted, the capacitance will changed, so the charge and energy will also get changed. But electric field did not changed as it is only depends on the battery voltage and the plate separation.

Conclusion: Capacitance, Charge and Energy will be changed when a dielectric is inserted.