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Chapter 13: Problem 46

The magnitude of electric intensity at a distance \(x\) from a charge \(q\) is \(E\). An identical charge is placed at a distance \(2 x\) from it. Then the magnitude of the force it experiences is (A) \(q E\) (B) \(2 q E\) (C) \(\frac{q E}{2}\) (D) \(\frac{q E}{4}\)

Short Answer

Expert verified
The magnitude of the force experienced by the second charge is \(\frac{q E}{4}\), which corresponds to option (D).

Step by step solution

01

Write the formula for electric field intensity

The electric field intensity (E) at a distance (x) from a charge (q) can be given by the formula: \(E = \frac{kq}{x^2}\), where \(k\) is Coulomb's constant, approximately \(8.99 脳 10^9 N路m^2/C^2\).
02

Calculate the force between the two charges

To find the force between two charges separated by distance, we use Coulomb's law: \(F = \frac{kq1q2}{r^2}\). In our case, both charges are equal (\(q1 = q2 = q\)), and the separation distance is \(2x\). Therefore, the formula becomes: \(F = \frac{kq^2}{(2x)^2}\)
03

Substitute E in the equation for force

We will now substitute the electric intensity (E) into the equation for F, using the equation from Step 1: \(E = \frac{kq}{x^2}\) First, we solve \(kq\) in terms of E and x: \(E x^2 = kq \Rightarrow kq = E x^2\) Now, we replace \(kq\) in the equation for F: \(F = \frac{(Ex^2)q}{(2x)^2}\)
04

Simplify the equation and find the force

The equation for the force can be simplified as follows: \(F = \frac{(E x^2)q}{(4x^2)}\) Notice that \(x^2\) can be canceled from the numerator and the denominator: \(F = \frac{Eq}{4}\) Hence, the magnitude of force experienced by the second charge is: \(F = \frac{q E}{4}\) Therefore, the correct choice is (D) \(\frac{q E}{4}\).

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Most popular questions from this chapter

The electric potential at a point situated at a distance \(r\) on the axis of a short electric dipole of moment \(p\) will be \(1 / 4\left(\pi \varepsilon_{0}\right)\) times (A) \(p / r^{3}\) (B) \(p / r^{2}\) (C) \(p / r\) (D) None of the above

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q .\) The potential difference between the centres of the two rings is (A) \(\frac{q R}{4 \pi \varepsilon_{0} d^{2}}\) (B) \(\frac{q}{2 \pi \varepsilon_{0}}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right]\) (C) Zero (D) \(\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right]\)

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 \(C\).

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.

Let \(u_{a}\) and \(u_{d}\) represent the energy density in air and in a dielectric, respectively, for the same field in both. Let \(K=\) dielectric constant. Then (A) \(u_{a}=u_{d}\) (B) \(u_{a}=K u_{d}\) (C) \(u_{d}=K u_{a}\) (D) \(u_{a}=(K-1) u_{d}\)
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