Question

The top left diagram below shows a 1-C positive charge placed in a uniform external electric field. The electric field in which the charge is placed is represented by field lines (drawn in red), and the resulting electric force on the charge is represented by the blue vector. Suppose the 1-C charge is removed, and is replaced, in turn, by each of the charges shown in the other three diagrams below. Assuming that the external electric field remains the same, draw force vectors on each of the other charges to indicate both the magnitude and direction of the electric force on the charge. (Scroll down for two more diagrams.)

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Answer 1

Concepts and reason

The concepts required to solve this problem are the electric force, charge, and electric field.

Initially, use the concepts of electric field and force in terms and draw the force diagram for the charge in the electric field.

Fundamentals

An electric field can be defined as a region around a charged particle in which a unit positive charge can experiences a force.

The expression of electric force vector is,

$\vec F = q\vec E$

Here, $q$ is the charge and $\vec E$ is the electric field.

The electric force can be defined as a force which exerts on a charged particle in the presence of the electric field.

(A)

Use the expression of electric force,

$\vec F = q\vec E$

Substitute $- 1{\rm{ C}}$ for $q$ in the equation.

$F = \left( { - 1{\rm{ C}}} \right)E$

The direction of the electric field is always from positive potential to negative potential, that is, higher potential to lower potential. So, when the negatively charged particle is placed in the electric field, then the negatively charged particle attracts towards the positive potential due to it has negative charge. And hence, the negatively charged particle -1 C moves in the direction opposite to the direction of the electric field.

Draw the electric force vector on the $- 1{\rm{ C}}$ charge in the given electric field as follows:

(B)

Use the following expression for the electric force.

$\vec F = q\vec E$

Substitute $- 2{\rm{ C}}$ for $q$ in the above equation.

$F = \left( { - 2{\rm{ C}}} \right)E$

The magnitude of electric field and direction of the electric field is same. The magnitude of charge is twice as compared to that of $1{\rm{ C}}$ . So, the magnitude of force is twice the force on $1{\rm{ C}}$ charge and the negative sign indicates the direction of force is opposite to the electric field.

Draw the electric force with twice the length as that of blue vector shown for $+ 1{\rm{ C}}$ charge and opposite in direction of blue vector.

(C)

Use the expression of electric force.

Substitute $+ 3{\rm{ C}}$ for $q$ in the equation $\vec F = q\vec E$ .

$F = \left( { + 3{\rm{ C}}} \right)E$

The magnitude of electric field and direction of the electric field is same. The magnitude of charge is thrice as compared to that of $1{\rm{ C}}$ . So, the magnitude of force is thrice as that on $1{\rm{ C}}$ charge but the positive sign indicates the direction of force is same as that of the electric field. Draw the electric force with thrice length as that of blue vector shown for $+ 1{\rm{ C}}$ charge and in same direction.

Ans: Part A

The diagram for the electric force vector of the negatively charge particle $- 1{\rm{ C}}$ is as shown,