Question

Find the force on the charge $\mathbf{+}\mathit{q}$in Fig. 3.14. (The $\mathit{x}\mathit{y}$plane is a grounded conductor.)

Short answer

Answer

The net force on q is $\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\left(\frac{{29}^2}{72d^2}\right)\mathit{z}$.

Step-by-step solution

Given data

Consider the given figure as shown below.

Here, from the given figure it is clear that $\mathit{x}\mathit{y}$plane is grounded conductor, thus potential is zero

Determine force

As $\mathbf{+}\mathit{q}$induces an equal and opposite charge of $\mathbf{-}\mathit{q}$at a distance of $\mathit{Z}\mathbf{=}\mathbf{-}\mathbf{3}\mathit{d}$and $\mathbf{-}\mathbf{2}\mathit{q}$induces $\mathbf{+}\mathbf{2}\mathit{q}$at distance of $\mathit{Z}\mathbf{=}\mathbf{-}\mathit{d}$.

Write the expression for force on $\mathbf{+}\mathit{q}$due to $\mathbf{-}\mathbf{2}\mathit{q}$,.

${\mathit{F}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}\left(-2q\right)}{{\left(2d\right)}^{\mathbf{2}}}\mathit{z}$ …… (1)

Write the expression for force on $\mathbf{+}\mathit{q}$due to $\mathbf{+}\mathbf{2}\mathit{q}$,.

${\mathit{F}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{q}\mathbf{)}}{{\mathbf{\left(}\mathbf{4}\mathbf{d}\mathbf{\right)}}^{\mathbf{2}}}\mathit{z}$ …… (2)

Write the expression for force on $\mathbf{+}\mathit{q}$due to $\mathbf{-}\mathit{q}$.

${\mathit{F}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{q}\mathbf{)}}{{\mathbf{\left(}\mathbf{6}\mathbf{d}\mathbf{\right)}}^{\mathbf{2}}}\mathit{z}$ …… (3)

Determine the net force

Write the expression for the net force on q .

$$\begin{gathered} \mathit{F}\mathbf{=}{\mathit{F}}_{\mathbf{1}}\mathbf{+}{\mathit{F}}_{\mathbf{2}}\mathbf{+}{\mathit{F}}_{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{-2q}{{\left(2d\right)}^2}+\frac{2q}{{\left(4d\right)}^2}+\frac{-q}{{\left(6d\right)}^2}\right]\mathit{z} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{-1}{2}+\frac{1}{8}-\frac{1}{36}\right]\mathit{z} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left(\frac{29q^2}{72d^2}\right)\mathit{z} \end{gathered}$$

Thus, the net force on q is $-\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\left(}\frac{\mathbf{29}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{72}{\mathbf{d}}^{\mathbf{2}}}\mathbf{\right)}\mathit{z}$.