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Charge \(Q\) is given a displacement \(\vec{r}=a \hat{i}+b \hat{j}\) in an electric field \(\vec{E}=E_{1} \hat{i}+E_{2} \hat{j}\). The work done is a. \(Q\left(E_{1} a+E_{2} b\right)\) b. \(Q \sqrt{\left(E_{1} a\right)^{2}+\left(E_{2} b\right)^{2}}\) c. \(Q\left(E_{1}+E_{2}\right) \sqrt{a^{2}+b^{2}}\) d. \(Q \sqrt{\left(E_{1}^{2}+E_{2}^{2}\right)^{2}} \sqrt{a^{2}+b^{2}}\)

Short Answer

Expert verified
The correct answer is (a) \( Q(E_1 a + E_2 b) \).

Step by step solution

01

Understand the Work Done Formula

The work done by a force when an object is displaced in a field is given by the dot product of the force vector and the displacement vector: \[ W = \vec{F} \cdot \vec{r} \] In the context of electric fields, the force due to the field on the charge \(Q\) is \( \vec{F} = Q \vec{E} \). The displacement is \( \vec{r} = a \hat{i} + b \hat{j} \). Therefore, the work done \(W\) can be expressed as: \[ W = Q \vec{E} \cdot \vec{r} \]
02

Compute the Dot Product

The dot product \( \vec{E} \cdot \vec{r} \) involves multiplying the respective components of \( \vec{E} = E_1 \hat{i} + E_2 \hat{j} \) and \( \vec{r} = a \hat{i} + b \hat{j} \) and summing them up: \[ \vec{E} \cdot \vec{r} = E_1 \cdot a + E_2 \cdot b \] Thus, \[ \vec{E} \cdot \vec{r} = E_1a + E_2b \]
03

Calculate the Work Done

Using the dot product result from Step 2, substitute it back into the work done formula: \[ W = Q \cdot (E_1a + E_2b) \] This simplifies directly to: \[ W = Q(E_1a + E_2b) \]
04

Identify the Correct Option

Comparing our result from Step 3 with the given options, we see that the expression matches option (a). Thus, the correct option is: \( Q(E_1a + E_2b) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric field
Electric fields are a fundamental concept in electromagnetism. They represent the force per unit charge exerted on a charged particle. Imagine them as invisible lines of force that influence charged objects in their vicinity. An electric field is described by vectors, which are characterized by both magnitude and direction.
  • The direction of the field is the direction that a positive test charge would accelerate if placed within the field.
  • The magnitude of the field is the force experienced per unit of charge.
Electric fields can be created by charged objects or can exist in empty space. Mathematically, an electric field \( \vec{E} \) can be expressed as the force \( \vec{F} \) divided by the charge \( Q \) as \( \vec{E} = \frac{\vec{F}}{Q} \). In our exercise, the electric field is given by \( \vec{E} = E_1 \hat{i} + E_2 \hat{j} \), which means it has components in the direction of the \( \hat{i} \) and \( \hat{j} \) unit vectors, typical of a two-dimensional plane.
Work done
In physics, 'work' refers to the energy transfer that occurs when a force moves an object over a distance. The work done by a force on an object can be calculated using the dot product of the force vector and the displacement vector. This is particularly relevant in the study of electric fields where the work done by an electric force on a charge can cause the charge to accelerate or move.
  • The formula is: \{ W = \vec{F} \cdot \vec{r} \}.
  • \( \vec{F} \) is the force vector while \( \vec{r} \) is the displacement vector.
  • When dealing with electric fields, the force \( \vec{F} \) experienced by a charge \( Q \) is given by \( Q \vec{E} \).
Thus, the work done \( W \) in moving a charge in an electric field \( \vec{E} \) is \{ W = Q \vec{E} \cdot \vec{r} \} which simplifies to \( W = Q(E_1a + E_2b) \). This tells us how much energy is imparted to the charge as it moves within the field.
Dot product
The dot product, also known as the scalar product, is a mathematical operation that takes two equal-length sequences of numbers—typically coordinate vectors—and returns a single number (a scalar). It is very useful in physics, particularly when dealing with vector quantities like force and displacement.
  • For two vectors \( \vec{A} = A_1 \hat{i} + A_2 \hat{j} \) and \( \vec{B} = B_1 \hat{i} + B_2 \hat{j} \), the dot product is given by \( \vec{A} \cdot \vec{B} = A_1B_1 + A_2B_2 \).
  • It results in a scalar, which simplifies calculations involving vectors.
  • The dot product is also related to the cosine of the angle between two vectors, though in component form like our exercise, it is straightforward to calculate.
In the context of the exercise, we found \( \vec{E} \cdot \vec{r} = E_1a + E_2b \) by applying the definition of the dot product to the electric field vector and the displacement vector. This fundamental operation allows us to determine the work done by the electric force, simplifying to \( W = Q(E_1a + E_2b) \).

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Most popular questions from this chapter

Which of the following is/are proportional to the inverse square of the distance \(x\) ? a. The potential at a distance \(x\) from an isolated point charge b. The electric field at a distance \(x\) from an isolated point charge c. The force per unit length between two thin, straight, infinitely long current carrying conductors, parallel to each other, separated by a distance \(x\). d. The electrostatic force between two large charged bodies kept at a small distance \(x\) apart.

The elcctric field lines are closer together near object \(A\) than they are near object \(B\). We can conclude a. the potential near \(A\) is greater than the potential near \(B\) b. the potential near \(A\) is less than the potential near \(B\) c. the potential near \(A\) is equal to the potential near \(B\) d. nothing about the relative potentials near \(A\) and \(B\)

Inside a hollow charged spherical conductor, the potential a. is constant b. varies directly as the distance from the center c. varies inversely as the distance from the center d. varies inversely as the square of the distance from the center

Two charged particles having charges 1 and \(-1 \mu \mathrm{C}\) and of mass \(50 \mathrm{gm}\) each arc held at rest while their separation is \(2 \mathrm{~m}\). Find the speed of the particles when their separation is \(1 \mathrm{~m}_{\text {. }}\) a. \(\frac{1}{5} \mathrm{~m} / \mathrm{s}\) b. \(\frac{3}{5} \mathrm{~m} / \mathrm{s}\) c. \(\frac{3}{10} \mathrm{~m} / \mathrm{s}\) d. \(\frac{2}{7} \mathrm{~m} / \mathrm{s}\)

A small positively charged sphere is placed inside a positively charged spherical shell. What happens if the inner sphere is connected with the outer shell by a conducting wire? a. The entire charge of inner sphere will be tansferred to outer shell and then both will be at same potential. b. The entire charge of inner sphere will be transferred to outer shell and then both will be at different potential. c. The entire charge of outer shell will be transferred to inner sphere and then both will be at same potential. d. Nothing can be predicted.

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