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Chapter 13: Problem 7

The electric field intensity at a point at a distance \(2 \mathrm{~m}\) from a charge \(q\) is \(E\). The amount of work done in bringing a charge of 2 coulomb from infinity to this point will be (A) \(2 E \mathrm{~J}\) (B) \(4 E \mathrm{~J}\) (C) \(\frac{E}{2} \mathrm{~J}\) (D) \(\frac{E}{4} \mathrm{~J}\)

Short Answer

Expert verified
The work done in bringing a charge of 2 coulombs from infinity to a point at a distance 2 meters from a charge q is (B) \(4 E \mathrm{~J}\). This is calculated by first finding the charge q using the electric field formula, then calculating the electric potential, and finally computing the work done using the test charge and the electric potential.

Step by step solution

01

Find the charge q

To find the charge q, we use the formula for the electric field: \(E = \frac{k * q}{r^2}\), where E is the electric field intensity, k is Coulomb's constant (\(8.9875 * 10^9 \frac{Nm^2}{C^2}\)), and r is the distance from the charge (2 meters). Rearranging this equation to solve for q, we get: \(q = \frac{E * r^2}{k}\)
02

Calculate the electric potential

To calculate the electric potential (V), we use the formula: \(V = \frac{k * q}{r}\) Substitute the value of q obtained in step 1: \(V = \frac{k * E * r^2}{k * r}\) The k terms cancel out, and we get: \(V = E * r\) Since r is 2 meters, the electric potential is: \(V = 2 * E\)
03

Compute the work done

Finally, we can calculate the amount of work done (W) to bring a 2 coulomb charge from infinity to a point 2 meters from charge q. The formula for calculating the work done is: \(W = q鈧� * V\) Substitute the value of V obtained in step 2 and q鈧� = 2 C: \(W = 2 * (2 * E)\) Therefore, the work done is: \(W = 4 * E\) The correct answer is (B) \(4 E \mathrm{~J}\).

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Most popular questions from this chapter

A charge \(q_{1}\) is placed at the centre of a spherical conducting shell of radius \(R\). Conducting shell has a total charge \(q_{2} .\) Electrostatic potential energy of the system (A) \(\frac{q_{1}^{2}+2 q_{1} q_{2}}{8 \pi \varepsilon_{0} R}\) (B) \(\frac{q_{2}^{2}+2 q_{1} q_{2}}{8 \pi \varepsilon_{0} R}\) (C) \(\frac{q_{1}^{2}+q_{1} q_{2}}{4 \pi \varepsilon_{0} R}\) (D) \(\frac{q_{2}^{2}+q_{1} q_{2}}{4 \pi \varepsilon_{0} R}\)

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) \(\frac{e^{2}}{m^{2}} g\)

The electric \(\vec{E}\) is given by \(\vec{E}=a \hat{i}+b \hat{j}\) (where \(a\) and \(b\) is constant and \(\hat{i}, \hat{j}\) are unit vector along \(x\) and \(y\) axis, respectively), the flux passing through a square area of side \(l\) and parallel to \(y=z\) plane is (A) \(b l^{2}\) (B) \(a l^{2}\) (C) \(\sqrt{\left(a^{2}+b^{2}\right)} l^{2}\) (D) \(\sqrt{\left(a^{2}-b^{2}\right)} l^{2}\)

Four charges equal to \(-Q\) are placed at the four corners of a square and a charge \(q\) is at its centre. If the system is in equilibrium, the value of \(q\) is (A) \(-\frac{Q}{4}(1+2 \sqrt{2})\) (B) \(\frac{Q}{4}(1+2 \sqrt{2})\) (C) \(-\frac{Q}{2}(1+2 \sqrt{2})\) (D) \(\frac{Q}{2}(1+2 \sqrt{2})\)

Charges \(Q_{1}\) and \(Q_{2}\) lie inside and outside, respectively, of a closed Gaussian surface \(S .\) Let \(E\) be the field at any point on \(S\) and \(\phi\) be the flux of \(E\) over \(S\), (A) If \(Q_{1}\) changes, both \(E\) and \(\phi\) will change. (B) If \(Q_{2}\) changes, \(E\) will change. (C) If \(Q_{1}=0\) and \(Q_{2} \neq 0\) then \(E \neq 0\) but \(\phi=0\). (D) If \(Q_{1} \neq 0\) and \(Q_{2}=0\) then \(E=0\) but \(\phi \neq 0\).
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