Question

If the total positive charge is Q = 1.62�10?6 C, what is the magnitude of the electric field caused by this charge at point P, a distance d = 1.53 m from the charge? (Part C figure)

Enter your answer numerically in newtons per coulomb.

Answer 1

Concepts and reason

The main concept required to solve this problem is the electric field and distance.

First, write the equation for the electric field. Use this equation and calculate the magnitude of the electric field caused by the given charge at point P at a given distance from the charge.

Fundamentals

The equation for the electric field at a point from a charge is,

$E = \frac{{kQ}}{{{r^2}}}$

Here, k is the electrostatic constant, Q is the charge and r is the distance of the charge from the point where the electric field is required to find.

The electric field can be defined as the region around a charged particle in which a unit positive charge can experiences a force. The electric field can be measured with the unit of N/C. The electric field can be denoted by a letter E. The direction of the electric field will be in the direction of the force between the charge and the unit positive charge.

The equation for the electric field due to the charge Q at a point P , at a distance d is,

$E = \frac{{kQ}}{{{d^2}}}$

Here, k is the Coulomb’s constant, Q is the charge and d is the distance of the point P, from the charge.

The equation for the electric field caused by the given charge at point P at a distance d is,

$E = \frac{{kQ}}{{{d^2}}}$

Substitute $8.99 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}$ for k, $1.62 \times {10^{ - 6}}{\rm{ C}}$ for Q, and $1.53{\rm{ m}}$ for d in above equation.

$\begin{array}{c}\\E = \frac{{\left( {8.99 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {1.62 \times {{10}^{ - 6}}{\rm{ C}}} \right)}}{{{{\left( {1.53{\rm{ m}}} \right)}^2}}}\\\\ = 6221{\rm{ N/C}}\\\end{array}$

Ans:

The magnitude of the electric field caused by the charge Q at point P is 6221 N/C.