/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Charge \(Q\) is given a displace... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 35

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) \(Q \sqrt{ \left.E_{1}^{2}+E_{2}^{2}\right)} \sqrt{a^{2}+b^{2}}\)

Short Answer

Expert verified
Given the options in the question, the work done is (A) \(Q(E_{1} a+E_{2} b)\)

Step by step solution

01

Express the Vectors

The electric field vector is \(\vec{E}=E_{1} i + E_{2} j\) and the displacement vector is \(\vec{r}=a i + b j\)
02

Compute the Dot Product

Now, we compute the dot product of \(\vec{E}\) and \(\vec{r}\). Dot product operation is performed by multiplying respective unit vectors and adding the results. Hence, \(\vec{E}.\vec{r}=E_{1} a + E_{2} b\)
03

Calculate Work Done

Finally, we substitute the dot product into the formula for work done, \(W = Q\vec{E}.\vec{r}\), which gives us the work done as \(W = Q(E_{1} a+E_{2} b)\)

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Most popular questions from this chapter

If electric field is given by \(\vec{E}=\left(\frac{1}{x^{2}}\right) \hat{i} \mathrm{~V} / \mathrm{m}\), the magnitude of potential difference between points \(x=10 \mathrm{~cm}\) and \(x=20 \mathrm{~cm}\) is (A) \(\mathbb{V}\) (B) \(2 \mathrm{~V}\) (C) \(5 \mathrm{~V}\) (D) \(10 \mathrm{~V}\)

Two point charges \(+8 q\) and \(-2 q\) are located at \(x=0\) and \(x=L\), respectively. The location of a point on the \(x\)-axis at which the net electric field due to these two point charges is zero (A) \(2 L\) (B) \(\frac{L}{4}\) (C) \(8 L\) (D) \(4 L\)

A position charge \(Q\) is uniformly distributed over the ring of radius \(R=1 \mathrm{~m}\), and potential at infinity is assumed to be zero. Column-I (A) Electric field at centre of ring is (B) Electric field at the axis of ring at a distance \(x=\sqrt{3} \mathrm{~m}\) from centre is (C) Electric potential at centre of ring is (D) Electric potential on axis of ring at a distance \(\sqrt{3} \mathrm{~m}\) from centre of ring is Column-II 1\. \(\frac{K Q}{2}\) 2\. \(K Q\) 3\. \(\frac{K Q \sqrt{3}}{8}\) 4\. Zero 5\. \(\frac{K Q \sqrt{2}}{7}\)

The electric potential \(V\) at any point \(x, y, z\) (all in metres) in space is given by \(V=4 x^{2}\) volts. The electric field (in \(\mathrm{V} / \mathrm{m}\) ) at the point ( \(1 \mathrm{~m}, 0,2 \mathrm{~m}\) ) (A) \(-8 \hat{i}\) (B) \(8 \hat{i}\) (C) \(-16 \hat{i}\) (D) \(8 \sqrt{5} \hat{i}\)

A thin spherical conducting shell of radius \(R\) has a charge \(q\). Another charge \(Q\) is placed at the centre of the shell. The electrostatic potential at a point \(P\) at a distance \(R / 2\) from the centre of the shell is (A) \(\frac{2 Q}{4 \pi \varepsilon_{0} R}\) (B) \(\frac{2 Q}{4 \pi \varepsilon_{0} R}-\frac{2 q}{4 \pi \varepsilon_{0} R}\) (C) \(\frac{2 Q}{4 \pi \varepsilon_{0} R}+\frac{q}{4 \pi \varepsilon_{0} R}\) (D) \(\frac{(q+Q)}{4 \pi \varepsilon_{0}} \frac{2}{R}\)
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