In the electrostatic field of a point charge q from point 1(Figure) we moved one and the same charge to points 2, 3, 4. Find work done on the charge during the movement in each case and compare them.
Explanation and Solution
The work done by an electrostatic field in moving a charge from one point to another depends only on the electric potential difference between the two points, since the electrostatic force is conservative.
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Given:
- We have a point charge \(q\) at the center of a circle.
- Points 1, 2, 3, and 4 lie on the same circle around the charge \(q\).
Step 1: Understand the electric potential
The electric potential \(V\) at any point on a circle centered around the charge \(q\) is the same because:
\[
V = \frac{kq}{r}
\]
where \(r\) is the radius of the circle. Since points 1, 2, 3, and 4 are equidistant from \(q\), the potential at all these points is identical.
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Step 2: Work done in moving a charge
The work done \(W\) in moving a charge \(Q\) from one point to another in an electrostatic field is given by:
\[
W = Q (V_{\text{final}} - V_{\text{initial}})
\]
Since the potential \(V\) is the same at points 2, 3, and 4, the potential difference for each movement is zero:
\[
V_{\text{final}} = V_{\text{initial}}
\]
Thus, for movements from point 1 to points 2, 3, and 4, the work done:
\[
W_2 = W_3 = W_4 = 0
\]
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Final Comparison:
The work done \(W_2\), \(W_3\), and \(W_4\) are all equal. Hence:
\[
{W_2 = W_3 = W_4 = 0}
\]
This result arises because the electric field is conservative and the movement is along an equipotential surface.