Question

Let '\(W\)' joule be the work done to move an electric charge '\(q\)' coulomb from a place \(A\), where potential is \(-5\text{ volt}\) to another place \(B\) where potential is '\(V\)' volt. The value of '\(V\)' is

• A. \(Wq - 5\)
• B. \(\frac{q}{W} + 5\)
• C. \(W - \frac{5}{q}\)
• D. \(\frac{W}{q} - 5\)

Hint:

Remember: Work = charge × potential difference.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The work done in moving a charge between two points in an electric field is directly related to the potential difference between those two points.
Step 2: Key Formula or Approach:
The work done \( W \) by an external force to move a charge \( q \) from point \( A \) to point \( B \) is given by the equation:
\( W = q \cdot \Delta V = q(V_B - V_A) \)
where \( V_A \) is the potential at the starting point and \( V_B \) is the potential at the final point.
Step 3: Detailed Explanation:
We are given the following values:
Work done \( = W \)
Charge moved \( = q \)
Potential at initial place \( A \), \( V_A = -5\text{ V} \)
Potential at final place \( B \), \( V_B = V \)
Substitute these values into the formula:
\[ W = q(V - (-5)) \] Simplify the expression inside the parenthesis:
\[ W = q(V + 5) \] Now, we need to solve for the final potential \( V \).
Divide both sides by \( q \):
\[ \frac{W}{q} = V + 5 \] Subtract 5 from both sides to isolate \( V \):
\[ V = \frac{W}{q} - 5 \] Step 4: Final Answer:
The value of \( V \) is \( \frac{W}{q} - 5 \).