Question

The figure shows a thin plastic rod of length $L = 13.5 cm$ and uniform charge $43.6 fC$ .

(a) In terms of distance $d$ , find an expression for the electric potential at a point $P_{1}$ .

(b) Next, substitute variable $x$ for $d$ and find an expression for the magnitude of the component $E_{x}$ of the electric field at $P_{1} .$

(c) What is the direction of $E_{x}$ relative to the positive direction of the $x$ -axis?

(d) What is the value of $E_{x}$ at $P_{1}$ for $x = d = 6.60 cm ?$

(e) From the symmetry in figure determine $E_{y}$ at $P_{1}$ .

Accepted Answer

(a) Electric potential at a point $P_{1}$ is,

$d V = \frac{k d q}{r}$ .

The distance from the point $P_{1}$ is, $r = d + x$

$\Rightarrow d V = \frac{k d q}{d + x}$ .

Since, the charge of the plastic and linear charge density of the rod is,

$\Rightarrow d q = λd x$

$λ = \frac{Q}{L}$ .

After substituting the linear charge density, the electric potential at the point is,

$\Rightarrow d V = \frac{k (λd x)}{(d + x)}$

Since, integrating the above equation to find the total potential at point $P_{1}$ ,

$\Rightarrow \int d V = \int \frac{k λd x}{d + x}$

$\Rightarrow V = k λ \int_{0}^{L} \frac{d x}{d + x}$

$L = 13.5 cm$

$\Rightarrow V = k \frac{Q}{L} \ln (\frac{d + L}{d})$

$\Rightarrow V = k \frac{Q}{L} \ln (1 + \frac{L}{d})$ .

Hence, the expression for the electric potential at the point $P_{1}$ is, $k \frac{Q}{L} \ln (1 + \frac{L}{d})$ .

(b)

The expression for the $x$ component of the electric field for the point $P_{1}$ is,

$E_{x} = - \frac{d V}{d x}$

$\Rightarrow E_{x} = - \frac{d}{d x} (k \frac{Q}{L} \ln (\frac{x + L}{x})) \Rightarrow E_{x} = - k \frac{Q}{L} (\frac{x}{x + L}) [L (- x^{- 2})]$

$\Rightarrow E_{x} = \frac{k Q}{x (x + L)}$ .

Hence, the expression for the $x$ -component of electric field for the point $P_{1}$ is, $E_{x} = \frac{k Q}{x (x + L)}$ .

(c) It is along the $- x$ direction.

(d) Putting the values, $E_{x} = \frac{k Q}{x (x + L)} = \frac{9 \times 10^{9} \times 43.6 \times 10^{- 15}}{6.6 (6.6 + 13.5) \times 10^{- 4}} = 2.96 \times 10^{- 2} N C^{- 1}$ .

(e)

Since the $y$ -component of the electric field is zero at the point $P_{1}$ because the direction of electric field is along $x - axis$ , so that $E_{y} = 0$ .

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