Question

The displacement of a charge \(Q\) in the electric field \(E^{-}=e_{1} i \wedge+e_{2} j \wedge+e_{3} k \wedge\) is \(r^{-}=a i \wedge+b j \wedge\) The work done is \(\ldots \ldots\) (A) \(Q\left(e_{1}+e_{2}\right) \sqrt{\left(a^{2}+b^{2}\right)}\) (B) \(Q\left[\sqrt{ \left.\left(e_{1}^{2}+e_{2}^{2}\right)\right](a+b)}\right.\) (C) \(Q\left(a e_{1}+b e_{2}\right)\) (D) \(\left.Q \sqrt{[}\left(a e_{1}\right)^{2}+\left(b e_{2}\right)^{2}\right]\)

Short answer

The short answer is: (C) \(Q(a e_{1}+b e_{2})\)

Step-by-step solution

Identify the force on the charge in the electric field

Since the problem gives us the electric field \(E^{-} = e_{1} i\wedge + e_{2} j\wedge+e_{3} k\wedge\), we can find the force on the charge by multiplying the electric field by the charge, Q. Therefore, the force can be written as: \(F^{-} = Q(e_{1} i\wedge + e_{2} j\wedge+ e_{3} k\wedge)\)

Identify the given displacement

The problem gives us the displacement of the charge as \(r^{-} = a i\wedge + b j\wedge\).

Compute the dot product of the force and displacement

The work done on the charge is the dot product of the force and displacement: \(W = F^{-} \cdot r^{-}\) Substitute the force and displacement expressions: \(W = Q(e_{1} i\wedge + e_{2} j\wedge+ e_{3} k\wedge) \cdot (a i\wedge + b j\wedge)\) To find the dot product, we can multiply component by component and sum them: \(W = Q (a e_{1} (i\wedge \cdot i\wedge) + b e_{2} (j\wedge \cdot j\wedge))\) Since \(i\wedge \cdot i\wedge =1\) , \(j\wedge \cdot j\wedge =1\) and \(i \cdot j = j \cdot k = k \cdot i = 0\), we get: \(W = Q(a e_{1} + b e_{2})\)

Identify the correct answer

Since we found that the work done is given by \(W=Q(a e_{1}+b e_{2})\), the correct answer is choice (C).

Key concepts

Dot Product

In the context of physics and particularly in electric fields, the dot product (also known as the scalar product) is a mathematical operation that combines two vectors.
The result of this operation is a scalar, which in our case represents the work done by a force operating over a displacement.

• The formula for the dot product of two vectors \( extbf{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \( extbf{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is: \( extbf{A} \cdot \textbf{B} = a_1b_1 + a_2b_2 + a_3b_3\).
• This operation requires multiplying corresponding components and then summing all results.
• In this exercise, it's essential to remember that the dot product of perpendicular components is zero. Thus, only the components aligned along the same axes contribute to the final scalar.

If you are computing work done by a force on a charge in an electric field, the dot product helps because it incorporates both the magnitude and the direction of the vectors involved.

Electric Field

An electric field describes the force that a charge experiences per unit charge in the space surrounding a charged object.
It is characterized by both a magnitude and a direction, represented as a vector field.

• The electric field vector \(\textbf{E}\) can be expressed in terms of its components, typically along the x, y, and z axes, such as \(E = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}\).
• It plays a pivotal role in determining the force on a charge, since the force \(\textbf{F}\) experienced by a charge \(Q\) is calculated as \(\textbf{F} = Q \cdot \textbf{E}\).

When exploring electric fields, it's crucial to recognize that they influence any charged particle within the field's vicinity.
The field's direction indicates the direction a positive charge would be pushed if placed within the field, while its magnitude relates to the strength of that push.

Displacement of Charge

The displacement of a charge refers to the change in position of that charge within an electric field.
It is also a vector quantity, incorporating both direction and magnitude.

• In vector form, displacement \(\textbf{r}\) might be given as \(\textbf{r} = a \hat{i} + b \hat{j}\), indicating movement along the x and y axes.
• Displacement is necessary to compute work because work depends not just on the force but also on how far and in what direction the charge moves.

Understanding displacement is critical when calculating work done since work involves assessing the influence of force along a certain distance.
In practical application, consider that movement in directions orthogonal to the force does not contribute to the work, as evidenced by the concept of the dot product.