Chapter 17, Problem 29P

(a)

To determine

The electric potential along the axis of dipole closer to positive charge will be calculated.

Answer to Problem 29P

The answer is $3.6\times {10}^{-2}V$ .

Explanation of Solution

Given data:

Dipole moment $p=4.8\times {10}^{-30}C\cdot m$

Distance $r=1.1\times {10}^{-9}m$

Formula used:

It is known that: $V=\frac{kp\cos \theta }{r^2}$

Where,

  $k$ is the constant factor.

  $\text{p}$ is the dipole moment.

  $V$ is the potential.

  $\text{r}$ is the distance between two charges.

Calculation:

Applying the formula to calculate potential,

  $V=\frac{kp\cos \theta }{r^2}$

  $\therefore V=\frac{\left(8.99\times {10}^{9}N\cdot m/C^2\right)\left(4.8\times {10}^{-30}C\cdot m\right)\cos 0}{{\left(1.1\times {10}^{-9}m\right)}^2}$

  $\therefore V=3.6\times {10}^{-2}V$

Conclusion: The required potential is $3.6\times {10}^{-2}V$ .

(b)

To determine

The electric potential $45°$ above the axis of dipole closer to positive charge will be calculated.

Answer to Problem 29P

The answer is $2.5\times {10}^{-2}V$ .

Explanation of Solution

Given data:

Dipole moment $p=4.8\times {10}^{-30}C\cdot m$

Distance $r=1.1\times {10}^{-9}m$

Angle, $\text{θ=}45°$

Formula used:

It is known that: $V=\frac{kp\cos \theta }{r^2}$

Where,

  $k$ is the constant factor.

  $\text{p}$ is the dipole moment.

  $V$ is the potential.

  $\text{r}$ is the distance between two charges.

Calculation:

Applying the formula to calculate potential,

  $V=\frac{kp\cos \theta }{r^2}$

  $\therefore V=\frac{\left(8.99\times {10}^{9}N\cdot m/C^2\right)\left(4.8\times {10}^{-30}C\cdot m\right)\cos 45°}{{\left(1.1\times {10}^{-9}m\right)}^2}$

  $\therefore V=2.5\times {10}^{-2}V$

Conclusion: The required potential is $2.5\times {10}^{-2}V$ .

(c)

To determine

The electric potential $45°$ above the axis of dipole nearer to negative charge will be calculated.

Answer to Problem 29P

The answer is $-2.5\times {10}^{-2}V$ .

Explanation of Solution

Given data:

Dipole moment $p=4.8\times {10}^{-30}C\cdot m$

Distance $r=1.1\times {10}^{-9}m$

Angle, $\text{θ=}135°$

Formula used:

It is known that: $V=\frac{kp\cos \theta }{r^2}$

Where,

  $k$ is the constant factor.

  $\text{p}$ is the dipole moment.

  $V$ is the potential.

  $\text{r}$ is the distance between two charges.

Calculation:

Applying the formula to calculate potential,

  $V=\frac{kp\cos \theta }{r^2}$

  $\therefore V=\frac{\left(8.99\times {10}^{9}N\cdot m/C^2\right)\left(4.8\times {10}^{-30}C\cdot m\right)\cos 135°}{{\left(1.1\times {10}^{-9}m\right)}^2}$

  $\therefore V=-2.5\times {10}^{-2}V$

Conclusion: The required potential is $-2.5\times {10}^{-2}V$ .