Question

23.28:At a certain distance from a point charge, the potential and electric field magnitude due to that charge are 4.98 V and 12.0 V/m, respectively. (Take the potential to be zero at infinity.)

1.What is the distance to the point charge? (d= ? m)
2.What is the magnitude of the charge? (q= ? c)

Answer 1

Concepts and reason

The concept required to solve this problem are electric field due to a point charge and electric potential due to a point charge.

Calculate distance d, solve the expression for both electric field and electric potential.

Fundamentals

Electric field is a vector field and at a point is the measure of force experienced by unit positive charge placed at that point by some source charge.

Electric field E at a point d distance apart from charge ${q_1}$ is given as follows:

$\vec E = \frac{{k{q_1}}}{{{d^2}}}\hat r$

Here, k is the proportionality factor and possess a constant value for a medium.

Electric potential at a point is the amount of work done in bringing a unit positive charge from a reference point generally at infinity to a specific point in the electric field without producing any acceleration.

Electric potential at a point d distance apart from charge ${q_1}$ is given by following expression:

$V = \frac{{k{q_1}}}{d}$

The expression for source charge is given as follows:

${q_1} = \frac{{Vd}}{k}$

(a)

Consider a point at distance d from a point charge ${q_1}$ .

The expression for magnitude of electric field E due to a point charge is given as follows:

$E = \frac{{k{q_1}}}{{{d^2}}}$

The expression for electric potential is given by following expression:

$V = \frac{{k{q_1}}}{d}$

Substitute the expression for potential in formula of electric field as follows:

$\begin{array}{c}\\E = \left( {\frac{{k{q_1}}}{d}} \right)\frac{1}{d}\\\\ = V\left( {\frac{1}{d}} \right)\\\end{array}$

The relation between electric field and electric potential is,

$E = \frac{V}{d}$

Rearrange the equation for distance.

$d = \frac{V}{E}$

Substitute $\left( {12.0{\rm{ V}} \cdot {{\rm{m}}^{ - 1}}} \right)$ for E and $\left( {{\rm{4}}{\rm{.98 V}}} \right)$ for V as follows:

$\begin{array}{c}\\d = \frac{V}{E}\\\\ = \frac{{\left( {{\rm{4}}{\rm{.98 V}}} \right)}}{{\left( {12.0{\rm{ V}} \cdot {{\rm{m}}^{ - 1}}} \right)}}\\\\ = 0.415{\rm{ m}}\\\end{array}$

(b)

Find the magnitude of the source charge.

The expression for electric potential is given by following expression:

$V = \frac{{k{q_1}}}{d}$

Rearrange the expression for potential for the charge as follows:

${q_1} = \frac{{Vd}}{k}$

Substitute $\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)$ for k, $\left( {0.415{\rm{ m}}} \right)$ for d and $\left( {{\rm{4}}{\rm{.98 V}}} \right)$ for V as follows:

$\begin{array}{c}\\{q_1} = \frac{{\left( {{\rm{4}}{\rm{.98 V}}} \right)\left( {0.415{\rm{ m}}} \right)}}{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)}}\\\\ = 0.23 \times {10^{ - 9}}{\rm{ C}}\\\end{array}$

Ans: Part a

The distance of a point charge from field point is $\left( {0.415{\rm{ m}}} \right)$ .

Part b

The magnitude of source charge ${q_1}$ is $\left( {0.23 \times {{10}^{ - 9}}{\rm{ C}}} \right)$ .