Question

Two charges $+6\,\mu\text{C}$ and $-3\,\mu\text{C}$ are placed at points $(-2.7\text{ cm},0)$ and $(2.7\text{ cm},0)$ respectively in an external electric field of $1.8\times10^5r^{-2}\,\text{NC}^{-1}$, where $r$ is the distance of a charge from the origin. Then the net electrostatic energy of the system of the two charges is:

• A. $63\text{ J}$
• B. $17\text{ J}$
• C. $23\text{ J}$
• D. $3\text{ J}$

Hint:

Whenever charges are placed in an external electric field, always calculate: \[ U=qV \] for each individual charge first and then add the mutual interaction energy \[ U_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r_{12}}. \] Remember that the distance used in the external potential calculation is measured from the origin, whereas the distance used in interaction energy is the separation between the charges.

Answer 1

The Correct Option is B

Concept: The total electrostatic potential energy of a system of charges placed in an external electric field consists of two separate contributions:

• Potential energy of each charge due to the external electric field.

• Mutual interaction energy between the charges themselves.
Thus, for two charges $q_1$ and $q_2$, \[ U_{\text{total}} = q_1V(r_1)+q_2V(r_2) + \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r_{12}}, \] where \[ V(r)=\text{electric potential due to the external field}, \] and \[ r_{12}=\text{distance between the two charges}. \] Therefore, our first task is to determine the electric potential corresponding to the given electric field.

Step 1: Determine the electric potential associated with the given electric field. The electric field is given as \[ E(r)=1.8\times10^5r^{-2}. \] The relationship between electric field and electric potential is \[ E=-\frac{dV}{dr}. \] Hence, \[ V=-\int E\,dr. \] Substituting the given field, \[ V = -\int 1.8\times10^5r^{-2}\,dr. \] Since \[ \int r^{-2}dr=-r^{-1}, \] we obtain \[ V = -\left[ 1.8\times10^5(-r^{-1}) \right]. \] Therefore, \[ V(r) = \frac{1.8\times10^5}{r}. \] Thus, the electric potential at a distance $r$ from the origin is \[ V(r)=\frac{1.8\times10^5}{r}. \]

Step 2: Calculate the electric potential at the locations of both charges. The coordinates of the charges are \[ (-2.7\text{ cm},0) \] and \[ (2.7\text{ cm},0). \] Both charges are at the same distance from the origin: \[ r_1=r_2=2.7\text{ cm}. \] Converting into SI units, \[ r_1=r_2=0.027\text{ m}. \] Therefore, \[ V(r_1)=V(r_2) = \frac{1.8\times10^5}{0.027}. \] Writing \[ 0.027=27\times10^{-3}, \] we get \[ V = \frac{1.8\times10^5}{27\times10^{-3}} = \frac{1.8}{27}\times10^8. \] Since \[ \frac{1.8}{27} = \frac{1}{15}, \] therefore \[ V = \frac{10^8}{15} = 6.67\times10^6\text{ V}. \] Equivalently, \[ V = \frac{2}{3}\times10^7\text{ V}. \]

Step 3: Calculate the potential energy of the first charge in the external field. For \[ q_1=+6\mu\text{C} = 6\times10^{-6}\text{ C}, \] the potential energy is \[ U_1=q_1V. \] Thus, \[ U_1 = (6\times10^{-6}) \left( \frac{2}{3}\times10^7 \right). \] Simplifying, \[ U_1 = 4\times10^1. \] Hence, \[ U_1=40\text{ J}. \]

Step 4: Calculate the potential energy of the second charge in the external field. For \[ q_2=-3\mu\text{C} = -3\times10^{-6}\text{ C}, \] the potential energy is \[ U_2=q_2V. \] Substituting, \[ U_2 = (-3\times10^{-6}) \left( \frac{2}{3}\times10^7 \right). \] Therefore, \[ U_2=-20\text{ J}. \]

Step 5: Calculate the mutual interaction energy between the two charges. The separation between the charges is \[ r_{12} = 2.7-(-2.7). \] Thus, \[ r_{12}=5.4\text{ cm}. \] Converting into SI units, \[ r_{12}=0.054\text{ m}. \] The interaction energy is \[ U_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r_{12}}. \] Substituting the values, \[ U_{12} = \frac{9\times10^9 (6\times10^{-6}) (-3\times10^{-6})} {0.054}. \] Simplifying, \[ U_{12} = \frac{-162\times10^{-3}} {54\times10^{-3}}. \] Therefore, \[ U_{12}=-3\text{ J}. \] The negative sign indicates attraction between opposite charges.

Step 6: Calculate the total electrostatic energy of the system. The total energy is \[ U_{\text{total}} = U_1+U_2+U_{12}. \] Substituting the calculated values, \[ U_{\text{total}} = 40+(-20)+(-3). \] Hence, \[ U_{\text{total}} = 17\text{ J}. \]

Final Conclusion: The net electrostatic energy of the two-charge system is \[ \boxed{17\text{ J}}. \] Therefore, the correct answer is \[ \boxed{\text{(B) }17\text{ J}}. \]