Question:

The value of electric potential at a distance r from a point charge is

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Don't confuse electric field and electric potential. For a point charge: - **Field (E)** is a vector and \(\propto 1/r^2\). - **Potential (V)** is a scalar and \(\propto 1/r\). Remember: Field is the 'slope' (derivative) of potential, which reduces the power of \(r\) by one.
  • proportional to r
  • inversely proportional to r
  • proportional to r\(^2\)
  • inversely proportional to r\(^2\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Electric potential (\(V\)) at a point in an electric field is defined as the work done in bringing a unit positive charge from infinity to that point. We need to find how this potential varies with distance from a single point charge.
Step 2: Key Formula or Approach:
The electric potential \(V\) at a distance \(r\) from a source point charge \(q\) is given by the formula: \[ V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \] where \(\frac{1}{4\pi\epsilon_0}\) is Coulomb's constant (\(k\)). So, \(V = k\frac{q}{r}\).
Step 3: Detailed Explanation:
From the formula \(V = k\frac{q}{r}\), we can see that for a given charge \(q\), the potential \(V\) is inversely proportional to the distance \(r\). \[ V \propto \frac{1}{r} \] This means that as the distance \(r\) from the charge increases, the electric potential decreases.
Option (D), inversely proportional to \(r^2\), describes the electric field strength (\(E \propto 1/r^2\)), not the potential.
Step 4: Final Answer:
The electric potential due to a point charge is inversely proportional to the distance \(r\) from the charge. Therefore, option (B) is correct.
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