Question

If the total positive charge is Q= 1.62×10^-6 , and the magnitude of the electric field caused by this charge at point P is 6220 N/C at a distance d= 1.53m from the charge. Now find the magnitude of the force on an electron placed at point P. Recall that the charge on an electron has magnitude e=1.60x10^-19 C .
Enter your answer numerically in newtons.

Answer 1

Concepts and reason

The concept of electric force is required to solve the problem.

Initially, use the relation between the electric force and electric field. Finally, substitute the value in the expression of the electrostatic force and calculate the result.

Fundamentals

The electric force on a charged object is as follows:

$F = qE$

Here, q is the charge and E is the electric field.

The magnitude of the force on an electron placed at point P in the figure is obtained by substituting e for q in the equation $F = qE$ .

Thus,

$F = eE$

Here, e is the charge on an electron.

The force on an electron is due to the electric field of the source charge Q.

Substitute $1.6 \times {10^{ - 19}}{\rm{ C}}$ for e and 6220 N/C for E in the expression $F = eE$ .

$\begin{array}{c}\\F = \left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {6220{\rm{ N/C}}} \right)\\\\ = 9.952 \times {10^{ - 16}}{\rm{ N}}\\\end{array}$

Ans:

The magnitude of force on an electron placed at point P is equal to $9.952 \times {10^{ - 16}}{\rm{ N}}$ .