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Chapter 13: Problem 118

Work done by electric field, when particle moves from \(C_{A}\) to \(C_{B}\) is (A) \(-1.2 \mathrm{~J}\) (B) \(1.2 \mathrm{~J}\) (C) \(-3.6 \mathrm{~J}\) (D) \(3.6 \mathrm{~J}\)

Short Answer

Expert verified
The work done by the electric field is \(-1.2 \; \mathrm{J}\).

Step by step solution

01

Identifying known variables

From the given problem, we identify the known quantities. The charge of the particle \(q\) is \(6 \times 10^{-3} \; \mathrm{C}\) and the potential difference (or voltage) between points \(C_{A}\) and \(C_{ B}\) is \(200 \; \mathrm{V}\).
02

Applying the formula for work done

The formula for the work done \(W\) by an electric field is \(W = qV\), where \(q\) is the charge and \(V\) is the voltage or potential difference.
03

Calculating the work done

Substitute the identified quantities into the equation: \(W = (6 \times 10^{-3} \; \mathrm{C})(200 \; \mathrm{V}) = 1.2 \; \mathrm{J}\).
04

Determining the direction of the work done

Work done by the electric field is considered negative because the electric field does work to move a particle against its field, therefore the work done by the electric field is \(-1.2 \; \mathrm{J}\).

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Most popular questions from this chapter

Three charges \(Q,+q\), and \(+q\) are placed at the vertices of a right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero, if \(Q\) is equal to (A) \(\frac{-q}{1+\sqrt{2}}\) (B) \(\frac{-2 q}{2+\sqrt{2}}\) (C) \(-2 q\) (D) \(+q\)

The magnitude of electric field intensity \(E\) is such that, an electron of mass \(m\) and charge \(e\) placed in it would experience an electrical force equal to its weight is given by (A) \(m g e\) (B) \(\frac{m g}{e}\) (C) \(\frac{e}{m g}\) (D) \(\frac{e^{2}}{m^{2}} g\)

An electric dipole is placed at an angle of \(30^{\circ}\) to a non-uniform electric field. The dipole will experience (A) a translational force only in the direction of the field. (B) a translational force only in a direction normal to the direction of the field. (C) a torque as well as a translational force. (D) a torque only.

A uniform electric field \(E=E_{0}(\hat{i}+\hat{j})\) exists in the region. The potential difference \(\left(V_{Q}-V_{P}\right)\) between point \(P(0,0)\) and \(Q(a, 0)\) is (A) \(-E_{0} a\) (B) \(E_{0} \sqrt{2} a\) (C) \(+E_{0} a\) (D) \(-E_{0} \sqrt{2} a\)

Two identical particles each having charge \(q\) are very far apart. They are given velocity \(v_{0}\) parallel to each other such that initial perpendicular separation between them is \(d\). If the subsequent minimum separation between them is \(2 d\), find the initial velocity \(v_{0}\) and the loss in their total kinetic energies.
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