Question:
Work done in assembling two identical charges each having a charge 'q', separated by a distance $r$ is:
Work done in assembling two identical charges each having a charge 'q', separated by a distance $r$ is:
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The work done in assembling point charges is always equal to the electrostatic potential energy of the final configuration.
For two charges, it is simply Coulomb's potential energy formula: $U = \frac{k q_1 q_2}{r}$.
For two charges, it is simply Coulomb's potential energy formula: $U = \frac{k q_1 q_2}{r}$.
Updated On: Jul 22, 2026
- $\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r}$
- $\frac{1}{4\pi\varepsilon_0}\frac{2q^2}{r}$
- $\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2r}$
- $\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the work done to bring two identical point charges, each of magnitude $q$, from infinity to a separation distance $r$.
Step 2: Key Formula and Approach:
The work done by an external agent in assembling a system of point charges is equal to the change in electrostatic potential energy ($U$) of the system:
\[ W = U \] For a system of two point charges $q_1$ and $q_2$ separated by a distance $r$, the potential energy is:
\[ U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r} \]
Step 3: Detailed Explanation:
• Formulate the assembly process:
1. Bringing the first charge $q_1 = q$ from infinity to its position requires no work because there is no existing electric field ($W_1 = 0$).
2. Bringing the second charge $q_2 = q$ from infinity to a distance $r$ from $q_1$ requires work against the electric field of $q_1$:
\[ W_2 = q_2 V_1 \] where $V_1$ is the electric potential due to $q_1$ at distance $r$:
\[ V_1 = \frac{1}{4\pi\varepsilon_0} \frac{q}{r} \]
• Calculate the total work done ($W$):
\[ W = W_1 + W_2 = 0 + q \left(\frac{1}{4\pi\varepsilon_0} \frac{q}{r}\right) \] \[ W = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r} \]
Step 4: Final Answer:
The work done in assembling the charges is $\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r}$, which corresponds to Option (A).
The question asks for the work done to bring two identical point charges, each of magnitude $q$, from infinity to a separation distance $r$.
Step 2: Key Formula and Approach:
The work done by an external agent in assembling a system of point charges is equal to the change in electrostatic potential energy ($U$) of the system:
\[ W = U \] For a system of two point charges $q_1$ and $q_2$ separated by a distance $r$, the potential energy is:
\[ U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r} \]
Step 3: Detailed Explanation:
• Formulate the assembly process:
1. Bringing the first charge $q_1 = q$ from infinity to its position requires no work because there is no existing electric field ($W_1 = 0$).
2. Bringing the second charge $q_2 = q$ from infinity to a distance $r$ from $q_1$ requires work against the electric field of $q_1$:
\[ W_2 = q_2 V_1 \] where $V_1$ is the electric potential due to $q_1$ at distance $r$:
\[ V_1 = \frac{1}{4\pi\varepsilon_0} \frac{q}{r} \]
• Calculate the total work done ($W$):
\[ W = W_1 + W_2 = 0 + q \left(\frac{1}{4\pi\varepsilon_0} \frac{q}{r}\right) \] \[ W = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r} \]
Step 4: Final Answer:
The work done in assembling the charges is $\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r}$, which corresponds to Option (A).
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