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Chapter 13: Problem 9

A charged particle of mass \(m\) and charge \(q\) is released from rest in an electric field of constant magnitude \(E\). The kinetic energy of the particle after time \(t\) will be (A) \(\frac{2 E^{2} t^{2}}{m q}\) (B) \(\frac{E q^{2} m}{2 t^{2}}\) (C) \(\frac{E^{2} q^{2} t^{2}}{2 m}\) (D) \(\frac{E q m}{2 t}\)

Short Answer

Expert verified
The kinetic energy of the charged particle after time $t$ is given by \(\frac{E^{2} q^{2} t^{2}}{2 m}\).

Step by step solution

01

Determine the electric force on the particle

In the presence of an electric field 'E', a charged particle with charge 'q' will experience an electric force 'F' given by the equation: \(F = qE\)
02

Use Newton's second law to find the acceleration

Newton's second law states that the acceleration 'a' of an object is directly proportional to the net force acting on it and inversely proportional to its mass 'm'. Using this law, we can determine the acceleration of the particle: \(a = \frac{F}{m}\) Substitute the electric force from Step 1: \(a = \frac{qE}{m}\)
03

Use equation of motion to find velocity after time 't'

As the particle starts from rest, the equation of motion for its velocity 'v' after time 't' can be given by: \(v = at\) Substitute the acceleration from Step 2: \(v = \frac{qE}{m}t\)
04

Calculate the kinetic energy

The kinetic energy 'K' of the particle can be given by the equation: \(K = \frac{1}{2}mv^{2}\) Substitute the velocity from Step 3: \(K = \frac{1}{2}m\left(\frac{qE}{m}t\right)^{2}\)
05

Simplify the expression for kinetic energy

Now, simplify the expression for kinetic energy: \(K = \frac{1}{2}m\left(\frac{q^{2}E^{2}t^{2}}{m^{2}}\right)\) \(K = \frac{q^{2}E^{2}t^{2}}{2m}\) Comparing the simplified expression with the given options, we can see that the correct answer is: (C) \(\frac{E^{2} q^{2} t^{2}}{2 m}\)

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Most popular questions from this chapter

A point charge \(q\) is placed at origin. Let \(\vec{E}_{A}, \vec{E}_{B}\) and \(\vec{E}_{C}\) be the electric field at three points \(A(1,2,3)\), \(B(1,1,-1)\), and \(C(2,2,2)\) due to charge \(q\). Then (A) \(\vec{E}_{A} \perp \vec{E}_{B}\) (B) \(\vec{E}_{A} \| \vec{E}_{B}\) (C) \(\left|\vec{E}_{B}\right|=4\left|\vec{E}_{C}\right|\) (D) \(\left|\vec{E}_{B}\right|=16\left|\vec{E}_{C}\right|\)

A capacitor of capacitance \(\mathrm{C}\) is fully charged by a \(200 \mathrm{~V}\) supply. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat \(2.5 \times 10^{2} \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) and of mass \(0.1 \mathrm{~kg}\). If temperature of the block rises by \(0.4 \mathrm{~K}\). Find the value of \(C\).

Choose the correct statements from the following: (A) If the electric field is zero at a point, the electric potential must also be zero at that point. (B) If electric potential is constant in a given region of space, the electric field must be zero in that region. (C) Two different equipotential surfaces can never intersect. (D) Electrons move from a region of lower potential to a region of higher potential.

Assume that an electric field \(\vec{E}=30 x^{2} \hat{i}\) exists in space. Then the potential difference \(V_{A}-V_{O}\), where \(V_{O}\) is the potential at the origin and \(V_{A}\) the potential at \(x=2 \mathrm{~m}\) is (A) \(120 \mathrm{~J}\) (B) \(-120 \mathrm{~J}\) (C) \(-80 \mathrm{j}\) (D) \(80 \mathrm{~J}\)

A body has a charge of one coulomb. The number of excess (or lesser) electrons on it from its normal state will be (A) \(\infty\) (B) \(1.6 \times 10^{-19}\) (C) \(1.6 \times 10^{19}\) (D) \(6.25 \times 10^{18}\)
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